 What is the molar mass of water?  What % of the mass comes from hydrogen?  What % of the mass comes from oxygen?

Unit 8: Stoichiometry

Add up the atomic masses of all the atoms in the formula. 2 mol H x 1.01 g/mol = 2.02 g H 1 mol O x g/mol = g O H 2 O molar mass g/mol

Assume 1 mole of the compound Divide each element mass by total 2.02 g H g H 2 O g O g H 2 O x 100% = 11.2% H x 100% = 88.8% O

CH 3 Fe 2 (SO 4 ) 3 Then name them. 6 mol X 1.01 g/mol = 6.06 g + 2 mol X g/mol = g g/mol 2 mol X g/mol = g + 3 mol X g/mol = g +12 mol X g/mol = g g/mol Dicarbon hexahydride (ethane) Iron (III) Sulfate

Calculate what percent of the molar mass comes from each type of element.

CH g H out of g CH 3 CH 3 = 6.06/30.08 = % H g C’s out of g CH 3 CH 3 = 24.02/30.08 = % C

Fe 2 (SO 4 ) 3 Fe S O

 Pick up a knowledge organizer and a textbook  What is the % composition of ammonium phosphate?

Gen Chem Unit 8 - Stoichiometry

 Empirical means, ‘by experiment.’  The empirical formula is usually the simplest mole ratio of elements, such as 1H:1O or HO.  So the simplest mole ratio gives us the empirical formula.

1. Determine how many grams of each element are present in the sample. 2. Determine how many moles of each element are present. 3. Divide all the moles by the smallest number of moles to get the mole ratios. 4. Complete the formula using the mole ratios

A substance contains 2 g of hydrogen and 32 g of oxygen. 1. How many moles of each? 2. What is the ratio of moles H to moles O? 3. What is the empirical formula? H x O y (what are x and y?)

2 g H x 1 mole H = 2 moles H 1 g H 32 g O x 1 mole O = 2 mole O 16 g O Mole ratio = 2:2 or 1:1 Possible formulas = HO, H 2 O 2, H 3 O 3 Empirical formula = HO

What is the empirical formula for a substance with a mass composition of: g copper, 48.2 g sulfur, and 96.0 g oxygen

What is the empirical formula for a substance with a mass composition of: g copper, 48.2 g sulfur, and 96.0 g oxygen

What is the formula for a substance with a mass composition of: 92.3 g copper x 1 mole Cu = 1.5 mol Cu 63.5 g Cu 48.2 g sulfur x 1 mole S = 1.5 mol S 32.1 g S 96.0 g oxygen x 1 mole O = 6 mol O 16.0 g O

What is the formula for a substance with a mass composition of: 1.5 mol Cu = 1 mol Cu mol S = 1 mol S mol oxygen = 4 mol O 1.5 Empirical formula CuSO 4 Empirical formula Cu x S y O z

Now you try it. What is the empirical formula for a substance with a mass composition of: 36.8 g of oxygen and 13.8 g carbon?

36.8 g O x 1 mole O = 2.3 moles O g O 13.8 g C x 1 mole C = 1.15 mole C g C = = Empirical Formula: CO 2

The percent composition of a substance is 78.1% B and 21.9% H. What is the empirical formula for this substance?

Assume you have 100g of the substance g B x 1 mole B = 7.23 moles B 10.8 g B 21.9 g H x 1 mole H = 21.9 mole H 1.0 g H = = Empirical Formula: BH 3

What is the formula for a substance with a percent composition of: 26.56% potassium, 35.41% chromium, and the remainder oxygen?

100 g g g = g O g K x 1 mole K = moles K 39.1 g K g Cr x 1 mole Cr = 0.68 mole Cr g Cr g 0 x 1 mole O = 2.37 mole O g O = = =

 1 K : 1 Cr : 3.5 O  2 K : 2 Cr : 7 O  Cr 2 K 2 O 7

 Many salts absorb water out of the air to form a salt hydrate.  You have 10 g of MgSO 4 hydrate. After heating, it weighs 4.89 g. Why?  How many water molecules are trapped inside a MgSO 4 crystal? MgSO 4 ∙ ? H 2 O

 4.89 g MgSO 4 x 1 mol MgSO 4 = mol g MgSO 4  5.11 g H 2 O x 1 mol H 2 O = mol g H 2 O = = MgSO 4 : 7 H 2 O  MgSO 4 ∙ 7H 2 O