Chemical formulas and composition stoichiometry Stoichiometry

The word “stoichiometry” is derived from the Greek stoicheion, which means “element,” and metron, which means “measure.”

Stoichiometry describes the quantitative relationships among elements in compounds (composition stoichiometry) and among substances as they undergo chemical changes (reaction stoichiometry). Main division of chemical compounds: organic vs inorganic

Many of the molecules found in nature are organic compounds.

• Organic compounds contain C-C or C-H bonds or both, often in combination with nitrogen, oxygen, sulfur, and other elements. • All of the other compounds are inorganic compounds.

Name Formula Name Formula Name Formula

water H2O sulfur dioxide SO2 butane C4H10

hydrogen peroxide H2O2 sulfur trioxide SO3 pentane C5H12

hydrogen chloride* HCl carbon monoxide CO benzene C6H6

sulfuric acid H2SO4 carbon dioxide CO2 methanol (methyl alcohol) CH3OH

nitric acid HNO3 methane CH4 ethanol (ethyl alcohol) CH3CH2OH

acetic acid CH3COOH ethane C2H6 acetone CH3COCH3

ammonia NH3 propane C3H8 diethyl ether (ether) CH3CH2OCH2CH3 Chemical formulas The chemical formula gives the number of atoms of each type in the molecule. But this formula does not express the order in which the atoms in the molecules are bonded together.

For instance, the structural formula of propane is C3H8 and shows that the molecule is composed of 3 carbons and 8 hydrogens.

The structural formula shows the order in which atoms are connected. The lines connecting atomic symbols represent chemical bonds between atoms The structural formula of propane shows that the three C atoms are linked in a chain, with three H atoms bonded to each of the end C atoms and two H atoms bonded to the center C. H H H

H CCC H

H H H

Chemists sometimes write out the formula in a way to better convey the connectivity information, e.g., CH3CH2CH3, which is a longer representation of the propane chemical formula. Chemical formulas

Each molecule of acetic acid, CH3COOH, contains 2 carbon atoms, 4 hydrogen atoms, and 2 oxygens.

Writing it as CH3COOH (instead of C2H4O2) includes useful bonding and structural information.

H OH H C C O H Law of Definite Proportions

A compound can be decomposed by chemical means into simpler substances, always in the same ratio by mass.

E.g.

Water originates from the reaction between two H and one O. This leads to the ratio 1:8 which is the ratio of the masses of H and O in the molecule of H2O.

AWH = 1,0079 amu = 1 AWO = 15,9994 amu = 16

One oxygen is sixteen times the mass of a H. Hence, in the molecule of water

H2 O H2O 2 amu 16 amu FW = 18 1 : 8

Water is 11.1% hydrogen and 88.9% oxygen by mass. Law of Definite Proportions

Carbon dioxide (CO2) is 27.3% carbon and 72.7% oxygen by mass Calcium oxide (CaO) is 71.5% calcium and 28.5% oxygen by mass Calcium carbonate (CaCO3) is 40.1% calcium, 12.0% carbon, 47.9% oxygen by mass

Observations such as these on many pure compounds (as CaCO3) led Joseph-Louis Proust in1799 to the statement of the Law of Deﬁnite Proportions (also known as the Law of Constant Composition):

Different samples of a compound always contain the same elements in the same proportion by mass; this corresponds to atoms of these elements combined in ﬁxed numerical ratios. 60 CHAPTER 2 • C H EMICAL FORMULAS A ND CO M POSITIO N STO I CHI O METR Y Percent2-7 composition Percent Composition and chemical and Formulas formulas of Compounds ▶ AW 5 atomic weight (mass)If the chemicalIf the formula formula of ais compound known, itsis known,chemical its chemical composition composition can canbe expressedbe expressed as MW 5 molecular weight (mass) the mass percent of each element in the compound (percent composition). For example, 60 the massCHAP percentTER 2 • CofH EMI eachCA Lelement FORMULAS in AtheND CcompoundO M POSITIO N (STpercentO I CHI O MET compositionR Y ). one carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is As a check, we see that the the part divided by the whole times 100% (or simply parts per hundred), so we can rep- E.g. percentages add to 100%. 2-7resent Percent the percent Composition composition of carbonand Formulas dioxide as follows: of Compounds One CO2, contains one C atom and two O atoms. ▶ AW 5 atomic weight (mass) PercentageIf the formulais themass part of C adivided compound by the is known, wholeAW of itsCtimes chemical 100%, composition so12.0 we amucan representcan be expressed the as MW 5 molecular weight (mass) the%C mass5 percent of each3 100% element5 in the compound3 100% (percent5 composition).3 100% For5 example,27.3% C percent compositionmass of CO of 2carbon dioxideMW as of follows: CO2 44.0 amu one carbon dioxide molecule, CO2, contains one C atom and two O atoms. Percentage is mass of O 2 3 AW of O 2(16.0 amu) As a check, we see that the %Othe5 part divided by3 100%the whole5 times 100% 3(or100% simply5 parts per hundred),3 100% so we5 72.7%can rep- O percentages add to 100%. resentmass the of percent CO2 compositionMW of ofcarbon CO2 dioxide as follows:44.0 amu

One mole of CO (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O atoms mass of C 2 AW of C 12.0 amu %C 5 (32.0 g). We could3 100% therefore5 have used these3 100% masses5 in the preceding3 100% calculation.5 27.3% These C massnumbers of CO are 2the same as theMW ones ofused COi2only the units are44.0 different. amu In Example 2-12 we masswill of Obase our calculation2 3on AWone mole of O rather than one 2molecule.(16.0 amu) %O 5 3 100% 5 3 100% 5 3 100% 5 72.7% O Onemass mole of of CO CO2 2 (44.0 g) containsMW of COone2 mole of C atoms44.0 (12.0 amu g) and two moles of O atoms (32.0 g). One mole of CO2 (44.0 g) contains one mole of C atoms (12.0 g) and two moles of O atoms (32.0 g). We could therefore have used these masses in the preceding calculation. These numbersEXA areMP LEthe 2-12 same Percent as the Compositionones usedionly the units are different. In Example 2-12 we will base our calculation on one mole rather than one molecule. Calculate the percent composition by mass of HNO3. Plan We first calculate the mass of one mole as in Example 2-8. Then we express the mass of each element as a percent of the total. Solution EXAMPLE 2-12 Percent Composition The molar mass of HNO3 is calculated first.

Calculate the percent composition by mass of HNO3. Number of Mol of Atoms 3 Mass of One Mol of Atoms 5 Mass Due to Element Plan 1 3 H 5 1 3 1.0 g 5 1.0 g of H We first calculate the mass of one mole as in Example 2-8. Then we express the mass of 1 3 N 5 1 3 14.0 g 5 14.0 g of N each element as a percent of the total. ▶ When chemists use the % notation, 3 3 O 5 3 3 16.0 g 5 48.0 g of O they mean percent by mass unless Solution they specify otherwise. Mass of 1 mol of HNO3 5 63.0 g The molarNow, its mass percent of HNO composition3 is calculated is first. mass of H 1.0 g Number of Mol of% Atoms H 5 3 Mass of3 One100% Mol5 of Atoms3 100%55Mass1.6% Due H to Element mass of HNO3 63.0 g 1 3 H 5 1 3 1.0 g 5 1.0 g of H mass of N 14.0 g 1 3 N 5 1 % N 5 3 14.0 g 3 100% 5 3 100%5514.022.2% g of N N mass of HNO3 63.0 g ▶ When chemists use the % notation, 3 3 O 5 3 3 16.0 g 5 48.0 g of O they mean percent by mass unless mass of O 48.0 g % O 5 3 100% 5 3 100% 5 76.2% O Mass of 1 mol of HNO3 5 63.0 g they specify otherwise. mass of HNO3 63.0 g Now, its percent composition is T o t a l 5 100.0% mass of H 1.0 g You should% now H 5 work Exercise 62. 3 100% 5 3 100% 5 1.6% H mass of HNO3 63.0 g mass of N 14.0 g % N 5 3 100% 5 3 100% 5 22.2% N mass of HNO3 63.0 g

Nitric acid is 1.6%mass H, 22.2% of O N, and 76.2% 48.0O by g mass. All samples of pure HNO have % O 5 3 100% 5 3 100% 5 76.2% O 3 this composition, accordingmass of HNO to the3 Law of Definite63.0 Proportions.g Unless otherwise noted, all content on this page is © Cengage Learning. T o t a l 5 100.0%

Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressedYou content should does not materially now affect work the overall Exercise learning experience. 62. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

10663_02_ch02_p043-080.indd 60 12/7/12 3:02 PM

Nitric acid is 1.6% H, 22.2% N, and 76.2% O by mass. All samples of pure HNO3 have this composition, according to the Law of Definite Proportions. Unless otherwise noted, all content on this page is © Cengage Learning.

Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

10663_02_ch02_p043-080.indd 60 12/7/12 3:02 PM Percent composition and chemical formulas Sample exercise • Calculate the percent composition by mass of HNO3 We ﬁrst calculate the mass of one mole (molar mass). Then we express the mass of each element as a percent of the total. The molar mass of HNO3 is: Number of Mol of Atoms x Mass of One Mol of Atoms = Mass Due to Element 1x H =1 mol x 1,0 g =1,0 g of H 1x N =1 mol x 14,0 g = 14,0 g of N 3x O = 3 mol x 16,0 g = 48,0 g of O Mass of 1 mol of HNO Now, its percent composition is: •% of H = (1.0 g /63,0 g) 100% = 1,6 % •% of N = (14.0 g /63,0 g) 100% = 22,2 % •% of O = (48,0 g /63,0 g) 100% = 76,2 %

Remember The amount of substance that contains the mass in grams numerically equal to its formula weight in amu contains one mole of the substance. This is the molar mass and is numerically equal to the formula weight and has the units grams/mole (g/mol). Percent composition and chemical formulas

Practice exercise • Calculate the percent composition by mass of each of the following compounds Dopamine: C8H11NO2 Vitamin E: C29H50O2 Vanillin: C8H8O3 Derivation of Formulas from Composition The molecular formula indicates the numbers of atoms in a molecule.

The empirical formula is the smallest whole-number ratio of atoms present.

E.g. The empirical and molecular formulas for water are both H2O; for hydrogen peroxide, the empirical formula is HO, and the molecular formula is H2O2

Molecular Simplest formula formula

C2H4O2 CH2O C6H12O6 CH2O H3PO4 H3PO4 H2O H2O H2O2 HO Derivation of Formulas from Composition

One of the ﬁrst steps in characterizing a new compound is the determination of its percent composition.

A qualitative analysis is performed to determine which elements are present in the compound.

A quantitative analysis is performed to determine the amount of each element.

Once the percent composition of a compound is known, the empirical formula can be determined 62 DerivationCHAPTER 2 of• C FormulasH EMICAL FORMUL fromAS A ND CCompositionO M POSITIO N STO I CHI O METR Y m

▶ A simple and useful way to obtain SampleStep exercise 2: Now (Empirical we know that Formula) 100.0 g of the compound contains 1.56 mol of S atoms andmol x Fw whole-number ratios among several numbers follows: (a) Divide each • Compounds 3.12containing mol of OS andatoms. O Weare obtain serious a whole-number air pollutants ratio and betweenthey represent these numbers the major number by the smallest number, and cause of acidthat rain. gives A sample the ratio of of this atoms compound in the simplest contains formula. 50.1% sulfur and 49.9% oxygen by then, (b) if necessary, multiply all the mass. What is the empirical formula of1.56 the compound? 5 1.00 S resulting numbers by the smallest 1.56 whole number that will eliminate 1.Let’s consider 100.0 g of compound, which contains 50.1 g of S andSO 49.92 g of O. We calculate the fractions 3.12 number of moles of atoms of each. 5 2.00 O 1.56 50,1 g 49,9 g Youmol should of S =now work Exercise= 1,5654. mol of S mol of O = = 3,12 mol of O 32,1 g mol-1 16 g mol-1 2.We obtain a whole-number ratio between these numbers that gives the ratio of atoms in the empirical formula 1,56 = 1,00 S 1,56 SO2 The solution for Example3.12 2-13 can be set up in tabular form. = 2,00 O 1,56 ▶ The “Relative Mass of Element” column is proportional to the mass of Relative Relative Number Smallest Whole- each element in grams. With this Mass of of Atoms Divide by Number Ratio interpretation, the next column could Element Element (divide mass by AW) Smallest Number of Atoms be headed “Relative Number of Moles 50.1 1.56 S 50.1 5 1.56 5 1.00 S of Atoms.” Then the last column 32.1 1.56 SO would represent the smallest whole- 49.9 3.12 2 number ratios of moles of atoms. But O 49.9 5 3.12 5 2.00 O 16.0 1.56 because a mole is always the same number of items (atoms), that ratio is the same as the smallest whole- This tabular format provides a convenient way to solve simplest-formula problems, as the number ratio of atoms. next example illustrates.

EXAMPLE 2-14 Simplest Formula A 20.882-g sample of an ionic compound is found to contain 6.072 g of Na, 8.474 g of S, and 6.336 g of O. What is its simplest formula? Plan We reason as in Example 2-13, calculating the number of moles of each element and the ratio among them. Here we use the tabu- lar format that was introduced earlier. Solution Relative Relative Number Convert Fractions Smallest Whole- Mass of of Atoms Divide by to Whole Numbers Number Ratio Element Element (divide mass by AW) Smallest Number (multiply by integer) of Atoms

6.072 0.264 Na 6.072 5 0.264 5 1.00 1.00 3 2 5 2 Na 23.0 0.264

8.474 0.264 S 8.474 5 0.264 5 1.00 1.00 3 2 5 2 S Na S O 32.1 0.264 2 2 3 6.336 0.396 O 6.336 5 0.396 5 1.50 1.50 3 2 5 3 O 16.0 0.264

The ratio of atoms in the simplest formula must be a whole-number ratio (by definition). To convert the ratio 1:1:1.5 to a whole- number ratio, each number in the ratio was multiplied by 2, which gave the simplest formula Na2S2O3 . You should now work Exercise 56.

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10663_02_ch02_p043-080.indd 62 12/7/12 3:02 PM 62 CHAPTER 2 • C H EMICAL FORMULAS A ND CO M POSITIO N STO I CHI O METR Y

▶ A simple and useful way to obtain Step 2: Now we know that 100.0 g of the compound contains 1.56 mol of S atoms and whole-number ratios among several numbers follows: (a) Divide each 3.12 mol of O atoms. We obtain a whole-number ratio between these numbers number by the smallest number, and that gives the ratio of atoms in the simplest formula. then, (b) if necessary, multiply all the 1.56 5 1.00 S resulting numbers by the smallest 1.56 whole number that will eliminate SO2 fractions 3.12 5 2.00 O 1.56 You should now work Exercise 54.

Derivation of Formulas from Composition m Sample exercise (Empirical Formula) mol x Fw • 20.882 g of an ionic compound is found to contain 6.072 g of Na, 8.474 g of S, and 6.336 g of TheO. What solution is its for empirical Example formula? 2-13 can be set up in tabular form. ▶ The “Relative Mass of Element” column is proportional to the mass of Relative Relative Number Smallest Whole- each element in grams. With this Mass of of Atoms Divide by Number Ratio interpretation, the next column could Element Element (divide mass by AW) Smallest Number of Atoms be headed “Relative Number of Moles 6,07250.1 g 1.560,264 NaS 6,07250.1 g 5= 1.560,264 mol 5 1.00= 1,00 S 1,00x2 = 2,00 of Atoms.” Then the last column 23,0 32.1g mol-1 1.560,264 SO would represent the smallest whole- 49.9 3.12 2 number ratios of moles of atoms. But O 49.9 8,474 g5 3.12 0,2645 2.00 O S 8,474 g 16.0 = 0,264 mol 1.56 = 1,00 1,00x2 = 2,00 because a mole is always the same 32,0 g mol-1 0,264 number of items (atoms), that ratio is 6,336 g 0,396 the same as the smallest whole- ThisO tabular format6,336 gprovides a convenient= 0,396 waymol to solve simplest-formula= 1,50 1,50x2 problems, = 3,00 as the 16,0 g mol-1 0,264 number ratio of atoms. next example illustrates.

Empirical formula = Na2S2O3

EXAMPLE 2-14 Simplest FormulaN.B. The ratio of atoms in the empirical formula must be a whole-number ratio (by deﬁnition). To convert the ratio 1:1:1.5 to a whole- number ratio, each number in the ratio was multiplied by 2, which gave the A 20.882-g sample of an ionic compound is found to contain 6.072 g of Na, 8.474 g of S, and 6.336 g of O. What is its simplest empirical formula Na S O formula? 2 2 3 Plan We reason as in Example 2-13, calculating the number of moles of each element and the ratio among them. Here we use the tabu- lar format that was introduced earlier. Solution Relative Relative Number Convert Fractions Smallest Whole- Mass of of Atoms Divide by to Whole Numbers Number Ratio Element Element (divide mass by AW) Smallest Number (multiply by integer) of Atoms

6.072 0.264 Na 6.072 5 0.264 5 1.00 1.00 3 2 5 2 Na 23.0 0.264

8.474 0.264 S 8.474 5 0.264 5 1.00 1.00 3 2 5 2 S Na S O 32.1 0.264 2 2 3 6.336 0.396 O 6.336 5 0.396 5 1.50 1.50 3 2 5 3 O 16.0 0.264

The ratio of atoms in the simplest formula must be a whole-number ratio (by definition). To convert the ratio 1:1:1.5 to a whole- number ratio, each number in the ratio was multiplied by 2, which gave the simplest formula Na2S2O3 . You should now work Exercise 56.

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10663_02_ch02_p043-080.indd 62 12/7/12 3:02 PM Derivation of Formulas from Composition

Practice exercise • The hormone norepinephrine is released in the human body during stress and increases the body’s metabolic rate. Like many biochemical compounds, norepinephrine is composed of carbon, hydrogen, oxygen, and nitrogen. The percent composition of this hormone is 56.8% C, 6.56% H, 28.4% O, and 8.28% N. What is the empirical formula of norepinephrine? Derivation of Formulas from Composition Sample exercise (Percent composition) • A 0.1647 g sample of hydrocarbon is burned in a C-H combustion train to produce 0.4931 g of CO2 and 0.2691 g of H2O. Determine the masses of C and H in the sample and the percentages of these elements in this hydrocarbon.

1. With a proportion, we use the observed masses to determine the masses of C and H in the original sample. There is one mole of carbons, (aw=12.01), in each mole of CO2, 44.01 g; there are two moles of hydrogen atoms, (aw=2.016), in each mole of H2O, 18.02 g: 12.01 g/mol C mass of g of C = 0.4931 g CO2 x = 0.1346 g C element AW 44.01 g/mol CO2 mass of the = FW 2.016 g/mol H compound g of H = 0.2691 g H2O x = 0.0301 g H 18.02 g/mol H2O

2. Then we calculate the percentages by mass of each element:

% C = 0.1346 g C x 100% = 81.72% C 0.1647 g sample

% H = 0.0301 g C x 100% = 18.28%H 0.1647 g sample

Remember: Hydrocarbons are organic compounds composed entirely of hydrogen and carbon. Molecular Formulas from Empirical Formulas To determine the molecular formula of a substance, both its empirical formula and its formula weight must be known.

The molecular formula is a multiple of the empirical formula. E.g. Butane, C 4H10. The empirical formula for butane is C2H5, but the molecular formula contains twice as many atoms; that is, 2x(C2H5)=C4H10.

Benzene, C 6H6. The empirical formula for benzene is CH. the molecular formula has six times as many atoms; 6x(CH)=C6H6. The molecular formula for a compound is either the same as, or an integer multiple of, the (empirical) formula. molecular weight so n = empirical formula weigh A practical example: Combustion Analysis One technique for determining empirical formulas in the laboratory is combustion analysis, commonly used for organic compounds.

When a compound containing carbon and hydrogen is combusted, the carbon is converted to CO2 and the hydrogen is converted to H2O. The amounts of CO and H O produced are determined by measuring the mass increase in the CO 2 SECTION2 3.5 Empirical Formulas from Analyses 95 2 and H2O absorbers. From the masses of CO2 and H2O we can calculate the number of moles of C and H in the original Combustion Analysis sample and thereby the empirical formula. One technique for determining empirical formulas in the laboratory is Sample combusted, H2O and CO2 are trapped combustion analysis, commonly used for compounds containing princi- producing CO2 and H2O in separate absorbers pally carbon and hydrogen. When a compound containing carbon and hydrogen is completely Ǡ FIGURE 3.14 Sample combusted in an apparatus such as that shown in ,the O2 carbon is converted to CO2 and the hydrogen is converted to H2O. •(Section 3.2) The amounts of CO2 and H2O produced are deter- H2O absorber CO2 absorber Furnace mined by measuring the mass increase in the CO2 and H2O absorbers. From the masses of CO2 and H2O we can calculate the number of moles Mass︎Apparatus gained for combustion by each absorber analysis corresponds to mass of of C and H in the original sample and thereby the empirical formula. If CO2 or H2O produced a third element is present in the compound, its massIf a can third be element determined is present in the compound, its mass can be determined by subtracting the measured į FIGURE 3.14 Apparatus for masses of C and H from the original sample mass. by subtracting the measured masses of C and H from the original sam- combustion analysis. ple mass.

SAMPLE EXERCISE 3.15 Determining an Empirical Formula by Combustion Analysis Isopropyl alcohol, sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Determine the empirical formula of isopropyl alcohol. SOLUTION

Analyze We are told that isopropyl alcohol contains C, H, and O Plan We can use the mole concept to calculate grams of C in the CO2 atoms and given the quantities of CO2 and H2O produced when a and grams of H in the H2O. These masses are the masses of C and H given quantity of the alcohol is combusted. We must determine the in the alcohol before combustion. The mass of O in the compound empirical formula for isopropyl alcohol, a task that requires us to cal- equals the mass of the original sample minus the sum of the C and H culate the number of moles of C, H, and O in the sample. masses. Once we have the C, H, and O masses, we can proceed as in Sample Exercise 3.13.

Solve To calculate the mass of C from the measured mass of CO2,we first use the molar mass ofCO2,44.0 g mol, to convert grams of CO2 to moles of CO2.Because each CO2 molecule has only one C atom, there is 1 mol> of C atoms per mole of CO2 molecules. This fact allows us to convert moles of CO2 to moles of C. Finally, we use the molar mass of C, 12.0 g, to convert moles of C to 1 mol CO2 1 mol C 12.0 g C grams of C: Grams C = (0.561 g CO2) = 0.153 g C 44.0 g CO2 1 mol CO2 1 mol C

The calculation for determining H mass from H2O mass 1 mol H2O 2 mol H 1.01 g H is similar, although we must remember that there are 2 Grams H = (0.306 g H2O) = 0.0343 g H 18.0 g H O 1 mol H O 1 mol H mol of H atoms per 1 mol of H2O molecules: 2 2 ¢ ≤ ¢ ≤ ¢ ≤ The mass of the sample, 0.255 g, is the sum of the Mass of O = mass of sample - (mass of C + mass of H) masses of C, H, and O. Thus, the O mass is = 0.255 g - (0.153 g + 0.0343 g) = 0.068 g O ¢ ≤ ¢ ≤ ¢ ≤ The number of moles of C, H, and O in the sample is 1 mol C Moles C = (0.153 g C) = 0.0128 mol C therefore 12.0 g C

1 mol H Moles H = (0.0343 g H) = 0.0340 mol H 1.01 g H ¢ ≤ 1 mol O Moles O = (0.068 g O) = 0.0043 mol O 16.0 g O To find the empirical formula, we must compare the ¢ ≤ relative number of moles of each element in the sam- ple. We determine relative number of moles by divid- ing each of our calculated number of moles by the 0.0128 0.0340 0.0043 C: = 3.0 H: ¢ = 7.9≤ O: = 1.0 smallest number: 0.0043 0.0043 0.0043 The first two numbers are very close to the whole numbers 3 and 8,giving the empirical formula C3H8O. Law of Multiple Proportions

Two (or more) elements may form more than one compound (e.g. CO and CO2, NO and NO2, H2O and H2O2). The Law of Multiple Proportions states that

“When two elements, A and B, combine and form more than one compound, the ratio of the masses of element B, in each of the compounds, can be expressed by small whole numbers”

E.g.: SO2 and SO3 provide an example. In SO2, two moles of oxygen combine with one mole of sulfur atoms In SO3, three moles of oxygen combine with one mole of sulfur atoms. Thus the ratio of oxygen atoms in the two compounds compared to a given number of sulfur atoms is 2:3.

Many similar examples, such as CO and CO2 (1:2 oxygen ratio) and H2O and H2O2 (1:2 oxygen ratio), are known.

H2 O H2O 2 amu 16 amu FW = 18 1 : 8 Law of Multiple Proportions Sample exercise (Empirical Formula) • What is the ratio of the numbers of oxygen atoms that are combined with a given number of nitrogen atoms in the compounds N2O3 and NO?

1.To compare the number of oxygen atoms, we must have equal numbers of nitrogen atoms. Because NO has half as many nitrogen atoms in its formula relative to N2O3, we must multiply it by a factor of 2 to compare the two elements on the basis of an equal number of nitrogen atoms. Once we show the number of atoms of each element present, we can cancel out the equal amounts of nitrogen atoms, leaving the ratio of oxygen atoms.

Oxygen N2O3 2N 3O 3O 3 = = = = ratio 2 (NO) 2N 2O 2O 2

The oxygen ratio in the two compounds N2O3 and NO is 3:2 Law of Multiple Proportions

Practice exercise • Show that the compounds water, H2O, and hydrogen peroxide, H2O2, obey the Law of Multiple Proportions.

• Nitric oxide, NO, is produced in internal combustion engines. When NO comes in contact with air, it is quickly converted into nitrogen dioxide, NO2, a very poisonous, corrosive gas. 1. What mass of O is combined with 3.00 g of N in (a) NO and (b) NO2? 2. Show that NO and NO2 obey the Law of Multiple Proportions. Other Interpretations of Chemical Formulas Sample exercise (Empirical Formula) • Once we master the mole concept and the meaning of chemical formulas, we can use them in many ways. For example, what mass of chromium (Cr) is contained in 35.8 g of (NH4)2Cr2O7?

1.The formula tells us that each mole of (NH4)2Cr2O7 contains two moles of Cr atoms, so we ﬁrst ﬁnd the number of moles of (NH4)2Cr2O7. We know that the molar mass is 252.0 g/mol.

m moles of 35.8 g = = 0.142 mols mol x Fw (NH4)2Cr2O7 252.0 g/mol

2.Each mol of (NH4)2Cr2O7 contains 2 moles of Cr, so we convert the number of moles of (NH4)2Cr2O7 into the number of moles of Cr atoms it contains, using the proportion moles 2 mol Cr = 0.142 mol (NH4)2Cr2O7 x = 0.284 mols of Cr 1 mol (NH4)2Cr2O7 3.We then use the atomic weight of Cr to convert the number of moles of chromium atoms to the mass of chromium. g of Cr = 0.284 mol Cr x 51.99 g/mol Cr= 14.76 g Cr Other Interpretations of Chemical Formulas

Practice exercise • Mercury occurs as a sulﬁde ore called cinnabar, HgS.

How many grams of mercury are contained in 578 g of pure HgS?