Get out your notes and locate the following problem:

896 dL of CO2 gas contains how many atoms?

2.4e24

Day 4 10-3

Activity Series LabDay 4 10-3

Propose an activity series based on the following results:

A + BY AY + B

A + CY NO RXN

Day 5 10-4

A > B

C > A

C AB

CHAPTER QUIZ FRIDAY 10-7!!!

Day 5 10-4

Read pages 360-361 AND complete # 15 on page 361.

Day 4 10-3

Fe(s) + Pb(NO3)2(aq) Fe(NO3)2(aq) + Pb(s)

Ca(s) + 2H2O(l) H2(g) + Ca(OH)2(aq)

Cl2(aq) + 2NaI(aq) 2NaCl(aq) + I2(aq)

Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)

Percentage Composition

Mass of element Mass of compound

X (100) =

% element in compound

… tells how much an element contributes to the mass of the compound

Percentage Composition

H2O???

2 H: 2 * 1.0079 g = 2.0158 g

1 O: 1 * 15.999 g = 15.999 g

18.015 g

(2.0158 g / 18.015) X 100 = 11.190% H

(15.999 g / 18.015) X 100 = 88.810% O

Percentage CompositionAn unknown compound w/ a mass of 0.237 g is extracted from the roots of a plant. Decomposition of the sample produces 0.0948 g of C, 0.1264 g of O, and 0.0158 g of H. What is the % composition of the compound?

Day 6 10-5

0.112 kL of CO gas contains how many grams? How many atoms?

140 grams CO

6e24 atoms

Review (if needed) pages 325-327 AND complete #s 33, 34, 35, and 36 on pages 326 and 327.

Day 5 10-4

Empirical formula – smallest whole-number mole ratio for a compound

Empirical Formulas

To Calculate:1. Convert all elements involved to

moles

2. Divide all elements by the smallest # of moles

3. Obtain smallest whole #ed ratio

A compound contains 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula?

Empirical Formulas

1. Convert all eles. involved to moles

2. Divide all eles. by the smallest # of moles

3. Obtain smallest whole #ed ratio

Ca = 0.337 moles

O = 0.675 moles

H = .675 moles

Ca = 0.337 mols / 0.337 mols = 1

O = 0.675 mols / 0.337 mols = 2

H = 0.675 mols / 0.337 mols = 2

1 Ca : 2 O : 2 H CaO2H2Ca(OH)2

Calculating Empirical Formulas

In an unknown compound you find 4.04 g of N and 11.46 g O. Empirical formula?

67.2 L of CH4 gas = ___ grams

44.8 mg of solid Carbon = ___ L

Day 1 10-6

Homework # 2 = now

Postlabs = Friday 10-7 (tomorrow)

Presentations = Wednesday 10-12

Chapter Quiz = Wednesday 10-12

Day 1 10-6

Empirical formula – smallest whole-number mole ratio for a compound

Empirical Formulas vs. Molecular Formulas

Molecular formula – actual # of atoms of each ele. in a molecular compound

Sometimes But not always! the same.

Molecular Formulas

To Calculate:

Compare the molar mass of the empirical formula to the molar mass of the molecular formula

Molecular formula – actual # of atoms of each ele. in a molecular compound

Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu

Empirical mass = 13.019 g/mol

x = 6

Molecular formula = C6H6

Determining Molecular Formulas

Determining Molecular Formulas

In an unknown compound you find 4.04 g of N and 11.46 g O. Empirical formula? This unknown compound has a molar mass of 108.0 g/mol. What is the molecular formula?

Empirical formula = N2O5

Empirical mass = 108.009 g/mol

Molecular formula = N2O5

A sample of a compound contains 6.0 g of C and 16 g of O. The total sample is 22 g. The molecular molar mass is 44 grams

Percentage Composition

Empirical formula

Molecular formula

Determining Molecular Formulas

Review section 10.3 (if needed) and complete #s 45-49

for # 49 the %s can be treated as grams and used to find moles…

Assignment due Tuesday 10-11

Calculate the mass of Cu produced?

Mass of beaker and Cu – mass of beaker

Calculating percent yield

Percent yield – a way to compare how much you “should” get to how much you actually got

Percent Yield = Actual

yieldTheoretic yieldX 100

A sample of a compound contains 6.0 g of C and 16 g of O. The total sample is 22 g.

Percentage Composition

Empirical formula

Determining Molecular Formulas

Pd. 6

- 2 molar mass convs.

- 2 Av.’s # convs.

- 2 Av.’s law convs.

- 2 multi-step convs.

- 1 percentage comp.

- 1 empirical formula

- 1 molecular formula

- 3 content ?s (no math)

14 questions

Mass of element Mass of compound

X (100) =% element in compound

H2O???2 H: 2 * 1.0079 g = ________ g

1 O: 1 * ________ g = _________ g

(__________________) X 100 = _______% H

(___________________) X 100 = _______% O

A compound contains 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula?

1. Convert all eles. involved to moles

2. Divide all eles. by the smallest # of moles

3. Obtain smallest whole #ed ratio

Ca = 0.337 moles

O = 0.675 moles

H = .675 moles

Ca = 0.337 mols / 0.337 mols = 1

O = 0.675 mols / 0.337 mols = 2

H = 0.675 mols / 0.337 mols = 2

1 Ca : 2 O : 2 H Ca(OH)2

13.5 g Ca

40 g Ca

1 mol Ca

10.8 g O

16 g O

1 mol O

0.675 g H

1 g H

1 mol H

Percentage Composition

Empirical formula

Molecular formula