Honors Chemistry POGIL: Hey,Bro - We’re Related! Unit 06 - Chemical Quantities Page 1 of 4
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Objective: Develop a series of steps for determining the mass percent composition or the empirical and molecular formulas of a compound. Identify the formula of a chemical compound from its composition. Calculate the mass percent composition from a molecular or empirical formula. Determine the empirical or molecular formula from mass percent composition or analysis information. Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp. 72−75.
Model I: Empirical Formulas and Percent Composition
H H O O H H C O H C C H C H C H C O H H H O H O
formaldehyde ace c acid lac c acid
molecular formula: CH2O molecular formula: C2H4O2 molecular formula: C3H6O3
empirical formula: CH2O empirical formula: CH2O empirical formula: CH2O
Concerning acetic acid:
# in molar mass mass contribution Element percentage of molar mass formula (g/mol) (g/mol)
C 2 12.01 2 × 12.01 = 24.02 (24.02 / 60.05) x 100 = 40.00
H 4 1.008 4 × 1.008 = 4.032 (4.032 / 60.05) x 100 = 6.714
O 2 16.00 2 × 16.00 = 32.00 (32.00 / 60.05) x 100 = 53.29
Molar mass = (24.02g + 4.02g + 32.00g) = 60.05g
Therefore, ace c acid’s percent composi on is 40.00% C (carbon), 6.714% H (hydrogen), and 53.29% O (oxygen).
1. In a manner similar to what was presented in the Model, above, determine the mass percent composition of lactic acid.
Revised 2017-2018 LCA Honors Chemistry POGIL: Hey,Bro - We’re Related! Unit 06 - Chemical Quantities Page 2 of 4 2. Compare the percent compositions of acetic acid (in the Model) and lactic acid (question 1). (a) What can you conclude about the empirical formulas of compounds with the same percent composition?
(b) Without doing any calculations, what is the percent composition of formaldehyde?
3. If you know a compound’s percent composition, do you automatically know what the compound’s molecular formula is? If you do, explain how. If you do not, what is the missing piece of information that would be used to determine the mol- ecular formula?
4. What is the molar mass of CH2O (the empirical formula of formaldehyde, acetic acid, and lactic acid)? Work . . .
5. (a) What is the molar mass of acetic acid divided by the molar mass of CH2O equal to?
(b) If you multiply the subscripts of CH2O (the empirical formula of acetic acid) by the answer to (a) to obtain a “new” formula, what formula do you get? To which compound does this formula belong?
6. (a) What is the molar mass of lactic acid divided by the molar mass of CH2O equal to?
(b) If you multiply the stoichiometric subscripts of CH2O by the answer (a) to obtain a “new” formula, what formula do you get? To which compound does this formula belong?
Model II: Finding the Empirical Formula Acetic acid is the active ingredient in vinegar. A 157.5 g sample of acetic acid is broken down into its elements through chemical analysis. The chemical analysis shows that there are 63.00g carbon, 83.93g oxygen, and 10.57g hydrogen.
How do you use this information to determine the empirical formula of acetic acid? Methodology Example Step 1: Substances & Masses: Identify and record the masses of the substance involved. 63.00g carbon 83.93g oxygen 10.57g hydrogen
Revised 2017-2018 LCA Honors Chemistry POGIL: Hey,Bro - We’re Related! Unit 06 - Chemical Quantities Page 3 of 4 Step 2: Mass to Moles: Covert the masses into moles. 1mol C 63.00g C × 12.01g C = 5.246mol C 1mol O 83.93g O × 16.00g O = 5.246mol O 1mol H 10.57g H × 1.008g H = 10.49mol H
Step 3: Moles to Smallest Ratio: Determine the smallest whole number ratio of moles. You 5.246mol C ÷ 5.246mol = 1 do this by identifying the smallest number of moles and 5.246mol O ÷ 5.246mol = 1 dividing each mole value by the smallest value to determine the smallest ratio. 10.49mol H ÷ 5.246mol = 2 Step 4: Ratio to Empirical formula: Use the numbers in the whole number ratio as the C : O : H subscripts in the empirical. 1 : 1 : 2 !
C 1O1H2 = COH2
7. What feature related to composition, do all compounds with the same mass percent composition have? €
8. A sample of sodium bicarbonate (baking soda) was found to consist of 9.122g Na, 0.4000g H, 4.766g C and 19.04g O. Using the given experimental information, determine the empirical formula for sodium bicarbonate.
Model III: Finding the Molecular Formula While an empirical formula provides the smallest whole number ratio of atoms in a compound, a molecular formula provides the actual number of atoms in a molecule of the compound. Finding the percent by mass of a sample will allow you to de- termine the empirical formula but will not allow you to find the actual formula of the molecule. A molecular formula is a whole number multiple of the empirical formula. Look at the following examples of empirical and molecular formulae.
Revised 2017-2018 LCA Honors Chemistry POGIL: Hey,Bro - We’re Related! Unit 06 - Chemical Quantities Page 4 of 4 Empirical Empirical Molecular Molecular Formula formula Mass formula formula Mass Multiple
CH2O 30.03g C6H12O6 180.18g 6
CH2O 30.03g C5H10O5 150.15g 5
HO 17.01g H2O2 34.02g 2
SO2 64.06g S2O4 128.12g 2
We need to find the empirical formula for ace c acid. We know the empirical formula is CH2O. You are given an addi onal piece of informa on: the molar mass of your compound (ace c acid) is 60.05 g/mol.
(CH2O)X = (30.03g)X = 60.05 g x = 2 so mul ply everything by 2
molecular formula is 2 x CH2O = C2H4O2
(The actual formula for ace c acid is HC2H3O2 or HCH3COO, since it is about the H a ached to the acetate ion.
9. One of the chlorofluorocarbons, that is used in refrigerator compressors and contributes to destruction of ozone in the upper atmosphere, has a molar mass of 132.9 g/mol and a percent composition of 53.34% Cl, 28.59% F, & 18.07% C. Use the experimental data provided to determine the empirical and the molecular formulas of this CFC compound.
10. (a) An ionic substance is found to have the % composition by mass: 82.0% mercury, 4.9% carbon, and 13.1% oxygen. What is the empirical formula of the compound? (b) The molar mass of the compound is determined to be 489.2 g/mol. Determine the “molecular formula” of the compound must be. What is the correct unambiguous name of the compound?
Revised 2017-2018 LCA