DO NOW 3/5/14

What is the molar mass of water?

What % of the mass comes from hydrogen?

What % of the mass comes from oxygen?

% COMPOSITION (BY MASS)Unit 8: Stoichiometry

MOLAR MASS OF COMPOUNDS

Add up the atomic masses of all the atoms in the formula.

2 mol H x 1.01 g/mol = 2.02 g H1 mol O x 16.00 g/mol = 16.00 g OH2O molar mass 18.02 g/mol

MOLAR MASS OF COMPOUNDS

Assume 1 mole of the compound

Divide each element mass by total

2.02 g H18.02 g H2O

16.00 g O18.02 g H2O

0.112 x 100% = 11.2% H

0.888 x 100% = 88.8% O

CALCULATE MOLAR MASS FOR

CH3CH3

Fe2(SO4)3

Then name them.

6 mol X 1.01 g/mol = 6.06 g+ 2 mol X 12.01 g/mol = 24.02 g 30.08 g/mol

2 mol X 55.85 g/mol = 111.7 g+ 3 mol X 32.06 g/mol = 96.18 g+12 mol X 16.00 g/mol = 192.00 g 399.88 g/mol

Dicarbon hexahydride(ethane)

Iron (III) Sulfate

PERCENT COMPOSITION

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

CALCULATE PERCENT COMPOSITION FOR

CH3CH3

6.06 g H out of 30.08 g CH3CH3

= 6.06/30.08 = 20.15 % H

24.02 g C’s out of 30.08 g CH3CH3

= 24.02/30.08 = 79.85 % C

CALCULATE PERCENT COMPOSITION FOR

Fe2(SO4)3

Fe

S

O

DO NOW;:

Pick up a knowledge organizer and a textbook

What is the % composition of ammonium phosphate?

EMPIRICAL FORMULASGen Chem Unit 8 - Stoichiometry

DETERMINING A COMPOUND FORMULA

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.

DETERMINING EMPIRICAL FORMULAS

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

DETERMINING A COMPOUND FORMULA

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?

HxOy (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 1:1Possible formulas = HO, H2O2, H3O3

Empirical formula = HO

What is the empirical formula for a substance with a mass

composition of:92. 3 g copper, 48.2 g sulfur,

and 96.0 g oxygen

What is the empirical formula for a substance with a mass

composition of:92. 3 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 Cu48.2 g sulfur x 1 mole S = 1.5 mol S

32.1 g S96.0 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 Cu1.5

1.5 mol S = 1 mol S1.5 6 mol oxygen = 4 mol O1.5

Empirical formula CuSO4

Empirical formula CuxSyOz

Now you try it. What is the empirical formula for a substance with a mass composition of:

36.8 g of oxygenand 13.8 g carbon?

36.8 g O x 1 mole O = 2.3 moles O

16.00 g O

13.8 g C x 1 mole C = 1.15 mole C

12.01 g C

= 2 1.15

= 11.15

Empirical Formula: CO2

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.

78.1 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

= 1 7.23

= 37.23

Empirical Formula: BH3

What is the formula for a substance with a percent

composition of:

26.56% potassium, 35.41% chromium, and the remainder

oxygen?

100 g- 26.56 g - 35.41 g = 38.03 g O

26.56 g K x 1 mole K = 0.679 moles K 39.1 g K

35.41 g Cr x 1 mole Cr = 0.68 mole Cr 52.00 g Cr

38.03 g 0 x 1 mole O = 2.37 mole O 16.00 g O

= 1 0.679

= 10.679 = 3.490.679

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

Cr2K2O7

Many salts absorb water out of the air to form a salt hydrate. You have 10 g of MgSO4 hydrate. After heating, it weighs 4.89 g. Why? How many water molecules are trapped inside a MgSO4 crystal?

MgSO4 ∙ ? H2O

4.89 g MgSO4 x 1 mol MgSO4 = 0.0406 mol

120.36 g MgSO4

5.11 g H2O x 1 mol H2O = 0.2836 mol

18.02 g H2O

= 1 0.0406

= 70.0406

1 MgSO4 : 7 H2O MgSO4 ∙7H2O