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Calculating Percent Composition

Calculating Percent Composition. Determination of Empirical and Molecular Formulas. Percent composition. Mass of the element x 100 = percent Mass of the compound. Percent Composition.

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Calculating Percent Composition

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  1. Calculating Percent Composition Determination of Empirical and Molecular Formulas

  2. Percent composition Mass of the element x 100 = percent Mass of the compound

  3. Percent Composition What is the percent carbon in C5H8NO4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? Mass of carbon x 100 = percent Mass of the glutamic acid

  4. C5H8NO4 • 8.22 %C b) 24.3 %C • 41.1 %C Find the mass of five carbon atoms and the molar mass of the compound. The correct answer is C.

  5. Find the percent water in a hydrate: Mass of water x 100 = percent water Mass of compound

  6. Find the percent of water in cobalt(II) chloride hexahydrate • Write the formula: CoCl2. 6 H2O 2) Calculate the mass of water in the compound and calculate the molar mass of the compound. 3) Divide and multiply by 100

  7. The percent of water in cobalt(II) chloride hexahydrate CoCl2. 6 H2O 6(18.02) x 100 = 45.4% 237.95

  8. Types of Formulas • Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. • Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

  9. Chemical Formulas of Compounds • Formulas give the relative numbers of atoms of each element in the compound- always a whole number ratio (the law of definite proportions). • If we can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

  10. Molecular vs empirical formulas • Ionic compounds are always empirical formulas. They are always reduced. Example: KI, MgBr2 Molecular formulas represent the atoms that the molecule is made of. They may or may not be reduced. Example: N2O4 is different than NO2.

  11. Which are molecular formulas and which are empirical? Which are both? Which are ionic? Which are covalent? H2O C6H6 NaCl MgSO4 (NH4)2CO3 C2H5OH C4H10

  12. Which are molecular formulas and which are empirical? Which are both? Which are ionic? Which are covalent? H2O both covalent compound C6H6 molecular covalent compound NaCl empirical ionic compound MgSO4 empirical ionic compound (NH4)2CO3 empirical ionic compound C2H5OH both covalent compound C4H10 molecular covalent compound

  13. To calculate an Empirical Formula 1. Determine the mass in grams of each element present, if necessary. 2. Calculate the number of moles of each element. 3. Divide each by the smallest number of moles to obtain the simplest whole number ratio. • If whole numbers are not obtained* in step 3, multiply through by the smallest number that will give all whole numbers *Be careful! Do not round off numbers prematurely

  14. Determine the empirical formula for this substance if 2.40 grams of carbon and 6.40 grams of oxygen are in the compound. Convert the grams to moles moles of C = 2.40g of C = 0.199 moles of C 12.01 g/mole moles of O = 6.40 g = 0.400 moles of O 16.00 g/mole Find the simplest whole number ratio and write the formula: 0.199 moles C = 1 0.400 moles O = 2 CO2 0.199 0.199

  15. Empirical Formula from % Composition A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance? Consider a sample size of 100 grams 60.80 grams of Na 28.60 grams of B and 10.60 grams H Determine the number of moles of each Determine the simplest whole number ratio NaBH4

  16. What is the empirical formula of a compound with the percent composition of 65.2% arsenic and 34.8% oxygen by mass? 65.2 grams As and 34.8 grams of O Convert to moles: 65.2 grams As = 0.870 moles 34.8 grams O = 2.18 74.9 g/mole 16.0 g/mole Find the simplest whole number ratio: 0.870 = 1 2.18 = 2.51 0.870 0.870 Make the ratio into whole numbers: 1 x 2 = 2 2.51 x 2 = 5.02 The ratio is 2 to 5 As2O5

  17. Molecular Formula Molecular Mass Empirical Formula mass • Just divide the molecular mass by the empirical formula mass to find how many times bigger the molecular formula is compared to the empirical formula

  18. The empirical formula of a carbon and hydrogen compound is found to be CH. The molecular mass is 78 g/mole. Find the molecular formula. Molecular mass /empirical formula mass 78/13 = 6 C6H6

  19. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g nitrogen and 5.34 g oxygen. Determine the empirical formula for this substance. Convert grams to moles moles of N = 2.34g of N = 0.167 moles of N 14.01 g/mole moles of O = 5.34 g = 0.334 moles of O 16.00 g/mole Formula:

  20. Calculation of the Molecular Formula The compound has an empirical formula of NO2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance? Look: 92.0 / 46.0 = 2 Double the formula! N2O4

  21. The molecular formula of nicotine is 162.1 g/mole. It contains 74.0% carbon, 8.7% hydrogen, and 17.3 % nitrogen.Use the data to find the empirical formula.Use the data to find the molecular formula.

  22. Answers The empirical formula for nicotine is: C5H7N The molecular formula for nicotine is: C10H14N2

  23. An oxide of sulfur contains 60% by the mass of oxygen. Deduce the empirical formula. Convert the masses to moles: 60 grams = 3.750 moles O 40 grams = 1.247 moles S 15.999 32.065 Find the simplest whole number ratio of the moles Oxygen: Sulfur: 3.750 = 3.00 1.247 = 1.00 1.247 1.247 The empirical formula = SO3

  24. A compound of nickel was analyzed and show to have the following composition by mass: Ni: 37.9%, S: 20.7%, O 41.4%Deduce the empirical formula 37.9 = 0.644 moles Ni 20.7 = 0.646 moles S 41.4 = 2.587 moles O 58.693 32.065 15.999 Find the simplest whole number ratio: 0.644 = 1.00 0.646 = 1.00 2.587 = 4.00 0.644 0.644 0.644 NiSO4

  25. The percent composition of a CFC is 17.8% Carbon, 1.5% hydrogen, 52.6% chlorine and some fluorine The Mr is 135 amuDetermine the empirical and molecular formulas. Assume the sample is 100.0 grams 17.8 grams carbon, 1.5 grams hydrogen, 52.6 grams chlorine and 28.1 grams fluorine

  26. Find the empirical formula: 17.8 g = 1.482 moles C 12.01 1.5 g = 1.488 moles H 1.0079 52.6 g = 1.484 moles Cl 35.453 28.1 g = 1.479 moles F 18.998 The ratio is 1:1:1:1 Empirical formula CHClF

  27. Find the molecular formula: The mass of the empirical formula, CHClF, is 67.559 Molecular formula = 135 = 2.06 Empirical formula 65.6 The molecular formula is C2H2Cl2F2

  28. Propane has the following composition by mass: 18.02 grams of carbon and 4.03 grams of hydrogen. The molecular mass is 44.09Find the empirical and molecular formulas. The empirical and molecular formula is C3H8

  29. Resource www.mccsc.edu/~nrapp/chemistrypowerpoint/Student%20Ch%2010%20Stoichiometry.ppt

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