Notes: Molar Mass, Percent Composition, Mole Calculations, and Empirical/Molecular Formulas In Chemistry, a Mole is: the unit that measures the amount of a substance - equals 6.022 x 10 23 particles of that substance Molar Mass is: the number of grams of a substance that is equal to 1 mole of that substance. -For elements, the molar mass is equal to the atomic mass on the periodic table. -For compounds, solve for molar mass by the following steps: i. Find the molar mass of each element in compound (the atomic mass in grams). ii. Multiply molar mass by the subscript of that specific element in the compound to get the weighted mass of that element. (If the is no subscript, then multiply by 1.) iii. Add the weighted masses of all elements in the compound together to get the molar mass of the entire compound. CALCULATING MOLAR MASS OF COMPOUNDS a. sodium hydroxide b. calcium cyanide c. magnesium phosphate d. iron (III) dichromate 57
Percent Composition is: The amount (in percent form) of each element in a compound. Chemists can calculate theoretical or experimental percent composition by mass. Experimental Lab analysis is done to chemically separate and measure the mass of each element in a compound. Then the mass of each element is calculated as a percentage of the total sample mass. Theoretical The chemical formula is used to determine the molar mass of the compound, and the mass of each element in the compound. Then the mass of each element is calculated as a percentage of the total sample mass. To find experimental percent composition, divide the mass of an element given by the total mass of the compound given and multiply by 100. Example: A compound with a mass of 48.72 g is found to contain 32.69 g of zinc and 16.03 g of sulfur. What is the percentage composition of the compound? Mass percent of Zn = Mass of Zn x 100 % Mass of Compound = 32.69 g x 100 % = 67.10 % 48.72 g Mass percent of S = Mass of S x 100 % Mass of Compound = 16.03 g x 100 % = 32.90 % 48.72 g Practice: A chemist took a 125 g sample of water and separated it into 13.5 g of hydrogen and 111.5 grams of oxygen. What is the experimental percent composition by mass of each element? To find theoretical percent composition, divide the weighted mass of the element by the total mass of the compound and multiply by 100. Example: CO 2 Molar Mass of C = 12.01g x 1 = 12.01 (weighted mass) 12.01g Molar Mass of O = 16.00g x 2 = 32.00 (weighted mass) % C = 12.01g / 44.01g x 100 = 27.29% Carbon %O = 32.00g / 44.01g x 100 = 72.71% Oxygen Practice: Find the percent composition of calcium cyanide Formula: +32.00g 44.01g (molar mass of C 58 % Ca = % C = % N =
Find the percent composition of magnesium phosphate Formula: % Mg = % P = % O = How many grams of magnesium are in 350 g of magnesium phosphate? What mass of magnesium phosphate contains 15 g of magnesium? TRY ON YOUR OWN 1. Find the molar mass of each compound below a. lithium carbonate b. calcium nitrate c. tin (IV) sulfate 2. Find the percent composition of tin (IV) sulfate % Sn = % S = % O = How many grams of tin are in 250 g of tin (IV) sulfate? 59
Mole Calculations How many is a mole? How heavy is a mole? 6.02 x 10 23 atoms/molecules/formula units = 1 mole molar mass in grams = 1 mole How much space does a mole occupy? for a gas at STP 22.4 L = 1 mole MOLE CONVERSIONS 1. Find the mass of 4.50 moles of diphosphorus pentoxide 2. How many moles is 250.0 g of copper (II) sulfate? 3. How many molecules are in 110 g of diphosphorus pentoxide? 5. How many atoms of Phosphorus are in 48g of diphosphorus pentoxide? 6. How many total atoms are in 48 g of diphosphorus pentoxide? 6. Convert 13.3 L of fluorine gas at STP to grams. 60
Make the following mole conversions ON YOUR OWN. STEP 1: Write the correct formula STEP 2: Determine the molar mass (if necessary) STEP 3: Use dimensional analysis to convert a. 575 g of sodium sulfate to moles b. 0.025 moles of phosphorus pentachloride to grams c. 14.0 L of nitrogen gas at STP to moles d. 15.0 g of iron (III) nitrate to formula units e. 8.02 x 10 23 molecules of carbon disulfide to grams f. 2.24 x 10 25 atoms of Chlorine to moles of Carbon tetrachloride g. How many atoms of oxygen are in 4.7 L of carbon dioxide gas at STP? 61
EMPIRICAL AND MOLECULAR FORMULAS Empirical formula: a. Assume you have 100g of the substance. b. Multiply percent of element s composition by 100g. This will give you the mass of that element in 100g of the substance. (If percent is not given, you will need to find it.) c. To find the # moles of each element in the compound, divide the mass of each element in the composition by that element s molar mass. d. Reduce to smallest ratio by dividing the number of moles of each substance by the smallest number. Then round to the nearest whole number. e. Assign subscripts. Example: 78.10% B, 21.90% H Moles of B in diborane = 100g x 0.7810 = 78.10g / 10.81g = 7.22 Moles of H in diborane = 100g x 0.2190 = 21.90g / 1.01g = 21.68 Empirical Formula = BH 3 7.22 mol B 1 mol B 21.68 mol H 3 mol H Molecular formula: a. Find the empirical formula first. b. Add all of the weighted masses of the elements in the empirical formula next. c. Divide the molar mass of the compound by the weighted mass from the empirical formula. Round to the nearest whole number. d. Multiply the subscripts for each element in the empirical formula by this number. This is the molecular formula. Example: The empirical formula for hydrogen peroxide is HO. The molar mass of hydrogen peroxide is 34.02g. Molar mass of H = 1.01g 1.01g Molar mass of O = 16.00g +16.00g 17.01g (mass of substance by empirical formula) 34.02g / 17.01 = 2 (Multiply the subscripts by this number. If there is no subscript, it is 1.) HO 2 x H 1 O 1 H 2 O 2 Determine the empirical formula of each compound below a. P 4 O 10 b. C 6 H 12 O 6 c. C 3 H 6 O 1. The empirical formula of a compound is CH 2 O. Its molar mass is 360 g/mol. Find its molecular formula. 2. The empirical formula of a compound is P 2 O 5. Its molar mass is 284 g/mol. Find its molecular formula. 62