Chemistry: Empirical & Molecular Formulas

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Chemistry: Empirical & Molecular Formulas Name______________________ 1) What is the empirical formula for C15H30O5H5? 2) A compound has an empirical formula of C2H4O3N. Its molar mass is 359 g. What is the molecular formula? 3) A compound contains 12.7 g iron & 7.29 g oxygen. Its molar mass is 87.85 g. What is its empirical formula? What is its molecular formula? 4) A compound contains 65.6% carbon, 9.45% hydrogen & 25.0% oxygen. Its molar mass is 1922.85 g. What is its empirical formula? What is its molecular formula? 5) A compound contains 92.2% carbon & 7.75% hydrogen. It has a molar mass of 118 g. What is its empirical formula? What is its molecular formula? 6) A compound has an empirical formula of C2H5O3N. Its molar mass is 454 g. What is its molecular formula? 7) What is the empirical formula for C20H32O4N4? 8) A compound contains 69.6% carbon, 9.76% hydrogen & 20.6% oxygen. Its molar mass is 465.72 g. What is its empirical formula? What is its molecular formula? 9) A compound contains 42.044 g carbon & 7.956 g hydrogen. Its molar mass is 114.26 g. What is its empirical formula? What is its molecular formula? 10) A compound contains 30.79 g copper & 19.31 g oxygen. Its molar mass is 207 g. What is its empirical formula? What is its molecular formula? 11) A compound contains 40.9% carbon, 4.58% hydrogen & 54.5% oxygen. Its molar mass is 176.1 g. What is its empirical formula? What is its molecular formula? 12) A compound has an empirical formula of CH2O & a gram molecular mass of 180.16 g. What is its molecular formula? 13) What is the empirical formula for C6H9 ? 14) A compound contains 34.241 g carbon & 5.759 g hydrogen. Its molar mass is 28.06 g. What is its empirical formula? What is its molecular formula? 15) A compound contains 47.377% carbon, 10.625% hydrogen & 42.078% oxygen. Its molar mass is 76.05 g. What is its empirical formula? What is its molecular formula? 16) A compound contains 43.638% phosphorus & 56.362% oxygen. Its molar mass is 283.88 g. What is its empirical formula? What is its molecular formula? 17) A compound contains 27.631 g carbon, 4.880 g hydrogen, 6.446 g nitrogen & 11.043 g oxygen. Its molar mass is 434.66 g. What is its empirical formula? What is its molecular formula? 18) A compound contains 39.640% carbon, 10.834% hydrogen, 23.121% nitrogen & 26.405% oxygen. Its molar mass is 121.19 g. What is its empirical formula? What is its molecular formula? Answers 1) C3H10OH 2) C8H16O12N4 3) FeO2 4) C105H180O30 5) C9H9 6) C10H25O15N5 7) C5H8ON 8) C27H45O6 9) C8H18 10) Cu2O5 11) C6H8O6 12) C6H12O6 13) C2H3 14) C2H4 15) C3H8O2 16) P4O10 17) C20H42N4O6 18) C4H13N2O2 .

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WS 4.2: Empirical Formula Problems
Name: Answer Key Period: ______ Date: __________ Chem B odds for HW and evens for review WS 4.2: Empirical Formula Problems Directions: Determine the empirical formula for the following substances. If a molecular formula cannot be reduced, write “cannot be reduced in the box for empirical formula” Molecular Formula Empirical Formula Molecular Formula Empirical Formula 1) C6H6 CH 2) C10H18 C5H9 3) C8H18 C4H9 4) C7H15O2 Not reducible 5) WO2 Not reducible 6) N2H4 NH2 7) C2H6O2 CH3O 8) P3F6 PF2 9) X39Y13 X3Y 10) IF5 Not reducible 11) C4H10 C2H5 12) N2O5 Not reducible 13) C6H12O6 CH2O 14) N2O4 NO2 15) P2F4 PF2 16) SF6 Not reducible 17) SO3 Not reducible 18) N2Cl4 NCl2 19) N2Br4 NBr2 20) P2O5 Not reducible Directions: Solve the following problems. You may need to use a separate sheet of paper. 1) What is the empirical formula for a compound which 3) Barry Um has a sample of a compound which weighs 200 contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of grams and contains only carbon, hydrogen, oxygen and oxygen? nitrogen. By analysis, he finds that it contains 97.56 grams of carbon, 4.878 g of hydrogen, 52.03 g of oxygen and 45.53 g of nitrogen. Find its empirical formula. 3 FeSO C5H3N2O2 2) A compound is found to contain 53.7 % iron and 46.27 % sulfur. Find its empirical formula. 4) What is the empirical formula for a compound which contains 80.3% zinc and the rest is oxygen? Fe2S3 ZnO 5) Rubbing alcohol was found to contain 60.0 % carbon, 13.4 9) Give the name and empirical formula of a compound % hydrogen, and the remaining mass was due to oxygen.
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Empirical and Molecular Formula Notes • the Empirical Formula of A
Empirical and Molecular Formula Notes The empirical formula of a compound shows the ratio of elements present in a compound The molecular formula of a compound shows how many atoms of each element are present in a molecule of a compound. o Example: the compound butene has a molecular formula of C4H8. The empirical formula of butene is CH2 because there is a 1:2 ratio of carbon atoms to hydrogen atoms. Finding the molecular formula when given the empirical formula The molecular formula is just a multiple of the empirical formula. Take a look at butene: o CH2 x n = C4H8 so n=4. What is the molar mass of butene? . If the empirical formula of a compound is CH2, then you can find the molar mass of the empirical formula: 12.0+ 2(1.0) = 14.0 g/mol . 14.0 g/mol x 4 = 56 g/mol What if I have a compound with an empirical formula of CH2 that has a molar mass of 84 g/mol. What is the molecular formula of this compound? o 14.0 g/mol x n = 84 g/mol so n = 6 o CH2 x 6 = C6H12 Try a few: 1. What is the empirical formula of C8H10? C6H12O6? 2. What is the molecular formula of a molecule that has an empirical formula of NO2 and a molar mass of 92 g/mol? 3. Consider a molecule that has an empirical formula of SO3 and a molecular formula of S3O9. What is the molar mass of the molecule? Calculating empirical formula from percent composition Answer this question: What is the empirical formula of a compound that contains 25.53% Mg and 74.47% Cl? Steps: 1.
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Jöns Jacob Berzelius 1 Jöns Jacob Berzelius
Jöns Jacob Berzelius 1 Jöns Jacob Berzelius J. J. Berzelius Jöns Jacob Berzelius (1779–1848) Born 20 August 1779 Väversunda, Östergötland, Sweden Died 7 August 1848 (aged 68) Stockholm, Sweden Nationality Sweden Fields Chemistry Institutions Karolinska Institute Alma mater Uppsala University Doctoral advisor Johann Afzelius Doctoral students James Finlay Weir Johnston Heinrich Rose Known for Atomic weights Chemical notation Silicon Selenium Thorium Cerium Notable awards Copley medal Jöns Jacob Berzelius (Swedish: [jœns ˌjɑːkɔb bæɹˈseːliɵs]; 20 August 1779 – 7 August 1848) was a Swedish chemist. He worked out the modern technique of chemical formula notation, and is together with John Dalton, Antoine Lavoisier, and Robert Boyle considered a father of modern chemistry.[1] He began his career as a physician but his researches in physical chemistry were of lasting significance in the development of the subject. He achieved much in later life as secretary of the Swedish Academy. He is known in Sweden as the Father of Swedish Chemistry. Berzelius Day is celebrated on 20 August in honour of him[2]. Biography Born at Väversunda in Östergötland in Sweden, Berzelius lost both his parents at an early age. Relatives in Linköping took care of him, and there he attended the school today known as Katedralskolan. He then enrolled at Uppsala University where he learned the profession of medical doctor from 1796 to 1801; Anders Gustaf Ekeberg, the discoverer of tantalum, taught him chemistry. He worked as an apprentice in a pharmacy and with a physician in the Medevi mineral springs. During this time he conducted analysis of the spring water. For his medical studies he examined the influence of galvanic current on several diseases and graduated as M.D.
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Empirical Formulas.Doc Homework Review for Class
6.2H Empirical Formulas.doc Homework Review for Class Calculating an Empirical Formula Given the following information calculate the empirical formula for the molecule. 1. 56.33 % Oxygen, 43.67 % Phosphorus Solution:P2O5 STEP 1. Change the percent to grams If 100 grams of this compound were present there would be: Phosphorus 43.67 grams Oxygen 56.33 grams STEP 2. Calculate the number of moles of each element. You will need the molar mass for each element. Atom Mass 1 Moles Molar _ Mass 43.67grams 1mole Phosphorus 1.4 moles of P 1 30.97grams Oxygen 56.33grams 1mole 3.5 moles of O 1 16.00grams STEP 3. Calculate the Subscripts for the Empirical Formula. 1.4 moles of Phosphorus 3.5 moles of Oxygen DIVIDE ALL NUMBERS BY THE SMALLEST NUMBER 1.4 / 1.4 = 1.00 Phosphorus 3.5 / 1.4 = 2.5 Oxygen To adjust these numbers to whole numbers multiply both by 2 to get 2 and 5. The 2 will go with the Phosphorus and the 5 goes with the O. P2O5 2. 32.38% Sodium, 22.65 % Sulfur, 44.99% Oxygen Solution:Na2SO4 STEP 1. Change the percent to grams If 100 grams of this compound were present there would be: Sodium 32.38 grams Sulfur 22.65 grams Oxygen 44.99 grams STEP 2. Calculate the number of moles of each element. You will need the molar mass for each element. Atom Mass Moles 32.38grams 1mole Sodium 1.4 moles of Na 1 23.00grams1 22.65grams Molar1mole_ Mass Sulfur .7 moles of S 1 32.06grams Oxygen 44.99grams 2.8 moles of O 1 1mole 16.00grams STEP 3.
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Empirical and Molecular Formulas
Empirical and Molecular Formulas • When the subscripts in a chemical formula represent the simplest ratio of the kinds of atoms in the compound, the formula is called an empirical formula. – Most ionic compounds are described with empirical formulas. • A molecular formula describes the actual numbers of atoms of each element in a molecule. Examples of Empirical and Molecular Formulas • Hydrogen peroxide – Molecular formula – H2O2 – Empirical formula – HO • Glucose – Molecular formula – C6H12O6 – Empirical formula – CH2O Calculating Empirical Formulas Step 1: If you are not given mass in grams for each element, convert the data you are given to grams of each element. • This may involve simple unit conversions. For example, you may Be given pounds or milligrams, which you convert to grams using unit analysis. • Sometimes you are given the percentage of each element in the compound. Assume that you have 100 g of compound, and change the numBers for the percentages to grams. Calculating Empirical Formulas Step 2: Convert grams of each element to moles By dividing By the atomic mass of the element. Step 3: Divide each mole value By the smallest and round your answers to whole numBers or common mixed fractions. Step 4: If you have a fraction after the last step, multiply all the mole values By the denominator of the fraction. Step 5: The resulting mole values correspond to the suBscripts in the empirical formula. Calculating Empirical Formulas Example Empirical Formula Calculation • An ionic compound used in the brewing industry to clean casks and vats and in the wine industry to kill undesirable yeasts and bacteria is composed of 35.172% potassium, 28.846% sulfur, and 35.982% oxygen.
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Chemistry 101 Spring 2002 Part 1
CHEMISTRY 101 SPRING 2002 EXAM 1 FORM A SECTIONS 501-509 DR. KEENEY-KENNICUTT Directions: (1) Put your name, S.I.D. number and signature on the free response part of the exam where indicated. (2) Each multiple choice question is actually 2 questions on your scanning sheet. If you are sure of an answer, put the same answer down for both questions for 5 pts. If you cannot decide between two answers, put one answer down for one question and the other answer down for the other question. If you get one correct you'll get half credit for 2.5 pts. If there is an ambiguous multiple choice question, use the last page to explain your answer. (3) Do NOT write on the envelope. (4) Bubble in OPTION A on the scanning sheet IF you want your grade posted. (5) When finished, put the free response answers in the envelope with the scanning sheet. You can keep the multiple choice part - the answers will be given to you as you leave. (6) There are a total of 29 questions (17 actual questions). PART 1 1&2. Which of the following name/symbol pairs is incorrect? (a) chromium/Cr (b) tungsten/W (c) potassium/K (d) tin/Ti (e) silver/Ag 3&4. The correct name for CH3COCH3 is: (a) benzene (b) acetone (c) methanol (d) ethanol (e) acetic acid 5&6. Two moles of Ca(CH3COO)2 contain: (a) 2 formula units of Ca(CH3COO)2 2+ (b) 2 Ca cations - (c) 4 moles of CH3COO anions (d) 2 x Avogadro's number of carbon atoms (e) 2 moles of hydrogen atoms ã Keeney-Kennicutt, 2002 A1 7&8.
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General Chemistry I CHEM-1030 Laboratory Experiment No. 3 the Empirical Formula of a Compound
General Chemistry I CHEM-1030 Laboratory Experiment No. 3 The Empirical Formula of a Compound Introduction An initial look at mass relationships in chemistry reveals little order or sense. Mass ratios of elements in a compound, while constant, do not immediately tell anything about a compound’s chemical formula. For instance, water always contains the same proportions of hydrogen (11.11% by mass) and oxygen (88.89% by mass) but these figures do not tell how the formula H2O is obtained. A chemical formula is a usually a whole number ratio showing the relative numbers of moles of each element present. A chemical formula can be determined from the mass of each element present in a sample of compound by taking into account their relative atomic masses. The unit of chemical quantity is not mass but the gram-molecular weight, or mole, which is the formula mass of a substance expressed in grams. As a result, a mole of any substance can be weighed out. Moreover, a mole of any substance contains the same number (Avogadro’s number, 6.022 x 1023) of formula units. For example, a mole of U weighs 238.029 g and contains Avogadro's number of uranium atoms. One mole of H2O weighs 18.015 g and contains Avogadro's number of H2O molecules. One mole of NaC1 weighs 58.443 g and contains Avogadro's number of NaC1 formula units, that is, NaCl ion pairs. In some cases, it is experimentally straightforward to find the mass of each element in a sample of compound.
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The Emergence of the Structure of the Molecule and the Art of Its Synthesis
Total Synthesis DOI: 10.1002/anie.200((will be filled in by the editorial staff)) The Emergence of the Structure of the Molecule and the Art of Its Synthesis K. C. Nicolaou* At the core of the science of chemistry lie the structure of the molecule, the art of its synthesis, and the design of function within it. These attributes elevate chemistry to an essential, indispensable, and powerful discipline whose impact on the life and materials sciences is paramount, undisputed, and expanding. Indeed, today the combination of structure, synthesis, and function is driving many scientific frontiers forward, including drug discovery and development, biology and biotechnology, materials science and nanotechnology, and molecular devices of all kinds. What connects structure and function is synthesis, whose flagship is total synthesis, the art of constructing the molecules of nature and their derivatives. The power of chemical synthesis at any given time is reflected and symbolized by the state of the art of total synthesis, and as such the condition and sophistication of the latter needs to be continuously nourished and advanced. In this review the understanding of the structure of the molecule, the emergence of organic synthesis, and the art of total synthesis are traced from the nineteenth century to the present day. 1. Introduction other sciences, technologies, and engineering, and how did it come to be so advanced and enabling? The power of chemistry is The celebration of Angewandte Chemie’s 125th anniversary in primarily derived from its ability to understand molecular structure, 2013 gives us the opportunity to reflect on both the past and the synthesize it, and build function within it through molecular design future of the central, and yet universal and ubiquitous, science of and synthesis.
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Composition of Substances and Solutions 129
Chapter 3 Composition of Substances and Solutions 129 Chapter 3 Composition of Substances and Solutions Figure 3.1 The water in a swimming pool is a complex mixture of substances whose relative amounts must be carefully maintained to ensure the health and comfort of people using the pool. (credit: modification of work by Vic Brincat) Chapter Outline 3.1 Formula Mass and the Mole Concept 3.2 Determining Empirical and Molecular Formulas 3.3 Molarity 3.4 Other Units for Solution Concentrations Introduction Swimming pools have long been a popular means of recreation, exercise, and physical therapy. Since it is impractical to refill large pools with fresh water on a frequent basis, pool water is regularly treated with chemicals to prevent the growth of harmful bacteria and algae. Proper pool maintenance requires regular additions of various chemical compounds in carefully measured amounts. For example, the relative amount of calcium ion, Ca2+, in the water should be maintained within certain limits to prevent eye irritation and avoid damage to the pool bed and plumbing. To maintain proper calcium levels, calcium cations are added to the water in the form of an ionic compound that also contains anions; thus, it is necessary to know both the relative amount of Ca2+ in the compound and the volume of water in the pool in order to achieve the proper calcium level. Quantitative aspects of the composition of substances (such as the calcium-containing compound) and mixtures (such as the pool water) are the subject of this chapter. 130 Chapter 3 Composition
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Chemical Formula Detective: Determining the Empirical Formula of a Hydrate
Bellevue College | CHEM& 161 Chemical Formula Detective: Determining the empirical formula of a hydrate Background Different elements can form chemical bonds to create compounds. For example, sodium and chlorine combine to form sodium chloride, NaCl. In the chemical formula NaCl, there is a 1:1 ratio of sodium ions:chloride ions. However, not all compounds form in a 1:1 ratio of their constituent elements. If they did, John Dalton would have been correct in 1803 when he proposed the chemical formula of water as HO∗. Of course, we now know that the correct chemical formula of water is H2O, in which there is a 2:1 ratio of hydrogen atoms to oxygen atoms. Since a mole is Avogadro’s number of atoms, H2O is also a 2:1 ratio of moles of hydrogen to moles of oxygen. Thus, the atom ratio is equivalent to the mole ratio (not a mass ratio) in a given chemical formula. As chemistry students you have learned how to predict chemical formulas of ionic compounds based on periodic trends and nomenclature rules, but it hasn’t always been that way. For hundreds of years, the chemical composition of compounds was studied experimentally, and the results generalized into the nomenclature rules used today. These rules allow the accurate prediction of chemical formulas for many ionic compounds without doing any experimentation. For example, the nomenclature rules can be used to correctly predict the formula of magnesium iodide as MgI2 rather than MgI. The curious student of chemistry will wonder how such a prediction could be verified by experimental means.
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Empirical Formula Worksheet
Empirical Formula When scientists discover a new compound they need to experiment to determine the chemical formula. They can do experiments where the percent composition of the new compound is found or they can measure how many grams of each element is present directly. Based on this data they can determines the lowest whole number ratio between particles using the chemistry counting unit=the mole. It is important to remember that empirical formulas are experimental formulas and will always be the lowest common ratio between the elements. CHO is an empirical formula. C6H12O6 is NOT an empirical formula; all of the elements can be divided by 6! Empirical Formula: The simplest ratio of the atoms present in a molecule. Circle the 6 empirical formulas in the list below: N2O2 CH4 H2O2 C6H12O6 MgSO4 NH3 H2O NO CaCO3 C12H22O11 The goal is for you to actually calculate an empirical formula when given either the % composition or the mass of each element present. Directions: Step 1: If you are given percent composition pretend the percents are grams and go to the next step. (Really- just change the % to a g for grams!!!) Step 2: Convert the mass of each element to moles of each element using the atomic masses from the periodic table. Step 3: Find the ratio of the moles of each element by dividing the number of moles of each by the smallest number of moles. (Your answers should be whole numbers and are used as the subscripts in the chemical formula) Step 4: Use the mole ratio to write the empirical formula Example: Find the empirical formula for the compound that contains 42.05 g of nitrogen and 95.95 g of oxygen.
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Ap Chemistry Empirical and Molecular Formula Worksheet
Ap Chemistry Empirical And Molecular Formula Worksheet Flowering and untrusty Ernesto jettisons while zoophobous Bubba expand her canakin abnormally and disunite foolishly. Sometimes empyreal Spud stampedes her ignobleness steeply, but unblended Billie decerebrates circumspectly or scamp blamefully. Mustier and whacking Mohan rarefy her Buchan jacket or war aerially. How many moles of glucose does steam contain? Stoichiometry, and neutrons. File Type PDF Answer For Pogil Mole Ratios Answer For Pogil Mole Ratios Getting the books answer for pogil mole ratios now is a type of inspiring means. There are related lessons on the molecular and chemistry empirical worksheet answers pdf and physical properties such as matter. Ap Chemistry Atomic Structure Multiple Choice Answers. Calculate the empirical formula. Which underlie these points are still accepted today? Plan: We immediately use dimensional analysis to lever the desired. Calc ulate the molecular weight of different compound. What move the chemical formula? This video shows a simple method that will crowd you have throw even an empirical formula in minutes! This page refresh the Oklahoma State University website provides an AP Chemistry test for students, etc. The interactive nature which our videos set us apart from database other curriculum sources. Students color based on the total expenditure of atoms in various compounds. Thus, consider the prick of the volumetric pipette with distilled water. This worksheet challenges chemistry students with nine questions that test their understanding of empirical formulas, can four determined using the data involve two different experiments. Later we cannot add acids. Physical change: require change leader does not speculate the chemical composition of much substance.
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