1. 294 g of sulfuric acid = ____ molecules

2. How many atoms of O?

1.8 * 1024 molecules

7.2 * 1024 atoms

Day 5 10-14

A chemical reaction produces 98.0 mL of sulfur dioxide gas at STP. What was the mass

(in grams) of the gas produced?

0.280 g SO2

896 dL of CO2 gas contains how many atoms?

2.4e24 molecules

7.2e24 atoms

Converting:Mass to moles

Moles to mass

Moles to atoms / molecules / particles

Atoms / molecules / particles to moles

Moles to liters

Liters to moles

Changing a substance

=

=

____________________________

______________________

=__________________________

= ________________

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 g) X 100 = 11.190% H

(15.999 g / 18.015 g) 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?

1) 45 L of hydrogen gas = how many grams?

Day 6 10-15

2) A sample of a compound contains 12.15 g of Mg and 19 g of F. What is the percent composition of this compound (SHOW WORK)?

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?

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

What result do you expect from a match test if ….

hydrogen is present?

carbon dioxide is present?

Day 1 10-16

67.2 L of CH4 gas = ___ grams

44.8 mg of solid Carbon = ___ L

Prelab # 5:

According to the reaction in # 1, for every ___ pieces of cupric chloride ___ pieces of copper are produced. As a result we should expect to produce ___ of a mole of cpper in the first reaction, which would be ___ grams of Cu. SHOW YOUR WORK!

A sample of a compound contains 3.003 g of C, 0.504 g of H, and 4.000 g of O. If the molar mass is 180.157 g/mol what is the molecular formula.

Empirical formula

Molecular Formula

Question of the Day

Day 2 10-17

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

Day 3 10-18

56,000 mL of O2 gas = how many molecules of O2? How many atoms of O?

- 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

- 2 content ?s (no math)

13 questions

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

Book #s 45-49 = NOW

Create-a-Test = momentarily

Presentations = Wednesday 10-12

Chapter Quiz = Wednesday 10-12

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

Desk clear and mentally prepare for Quest.

Quest

Group Quest

Presentations

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

Homework # 2 = now

Postlabs = Friday 10-7 (tomorrow)

Presentations = Wednesday 10-12

Chapter Quiz = Wednesday 10-12

Day 1 10-6

1. A mole is a _________, plus it relates to ______.

2. Molar mass of Be(NO3)2

3. 3.0 grams of Be(NO3)2 = ___ moles

4. = ___ molecules?