Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole...

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Page 1: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.
Page 2: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

StoichiometryBalancing Equations

Molecular and Empirical Formulas

Percent Composition

Mole Conversions

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Page 3: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Mole Conversion for 100 points

¿How many moles are in 28 grams of CO2?

• 28.0gCO2 x 1 mol CO2 = 0.636 mol CO2 44.01gCO2

Page 4: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Mole Conversion for 200 points

How many moles of magnesium is 3.01 x 1022 atoms of magnesium?

• 3.01e22 atom Mg x 1 mole Mg = 0.0499 mol Mg 6.022e23 atoms Mg

Page 5: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Mole Conversion for 300 points

Determine the volume, in liters, occupied by 0.030 moles of a gas at STP.

• 0.030 mol x 22.4 L = 0.67 L 1 mole

Page 6: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Mole Conversion for 400 points

How many oxygen molecules are in 3.36 L of oxygen gas at STP?

• 3.36 L O2 x 1 mole O2 x 6.022e23 molecules = 9.03e22 22.4 L O2 1 mole O2

Page 7: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Mole Conversion for 500 point

Find the mass in grams of 2.00 x 1023 molecules of F2.

• 2.00e23 mc F2 x 1 mole F2 x 38g F2 = 12.6g F2

6.022e23 mc F2 1 mole

Page 8: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Percent Composition for 100 points

Find the percent composition of CuBr2

Cu = 63.55 63.55(1) x 100 = 28.45%223.35

Br x2 =+ 159.8 79.9(2) x 100 = 71.55%223.35

223.35

Page 9: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Percent Composition for 200 points

Find the percent composition of NaOHNa = 22.99 (1)22.99 x 100 = 57.48%

40.00O = 16.00 (1)16.00 x 100 = 40.00%

40.00H = + 1.01 (1)1.01 x 100 = 2.53%

40.00 40.00

Page 10: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Percent Composition for 300 points

Find the percent composition of N2S2N x2 = 28.02 14.01(2) x 100 = 30.44%

92.04S x2 =+64.02 32.01(2) x 100 = 69.56%

92.04 92.04

Page 11: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Percent Composition for 400 points

Find the percent composition of KMnO4

K = 39.10 (1)39.10 x 100 = 24.74% 158.04

Mn = 54.94 (1)54.94 x 100 = 34.76% 158.04

O x4= + 64.00 (4)16.00 x 100 = 40.50% 158.04

158.04

Page 12: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Percent Composition for 500 points

Find the percent composition of Al2(SO4)3

Al x2= 53.96 (2)26.98 x 100 = 15.78% 341.99

S x3 = 96.03 (3)32.01 x 100 = 28.08% 341.99

O x12= + 192 (12)16.00 x 100 = 56.14% 341.99

341.99

Page 13: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Empirical and Molecular Formulas for 100 points

Find the empirical formula for a compound containing .783% Carbon, .196% Hydrogen,

and .521% Oxygen..783g C x 1 mol C = 0.0652 = 2

12.01g C 0.0326.196g H x 1 mol H = .194 = 6 =C2H6O

1.01g H 0.0326.521g O x 1 mol O = 0.0326 = 1

16.00g O 0.0326

Page 14: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Empirical and Molecular Formulas for 200 points

The empirical formula for a substance is CH2O. If its molar mass is 180, what is the molecular

formula

C = 12.01 H x2 = 2.02 180 = 6 = 6(CH2O) = C6H12O6 O = +16.00 30.03

30.03

Page 15: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Empirical and Molecular Formulas for 300 points

The empirical formula for a substance is C3H7NO3. If its molar mass is 105.11, what is the

molecular formula

C x3 = 36.03 H x7 = 7.07 N x1 = 14.01 105.11 = 1 = 1(C3H7NO3) = C3H7NO3 O x3 = +48.00 105.11

105.11

Page 16: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Empirical and Molecular Formulas for 400 points

Find the empirical formula for a compound containing 1.388g Carbon, .345g Hydrogen, and 1.850g Oxygen.

1.388g C x 1 mol C = 0.1156 = 1 12.01g C 0.1156

.345g H x 1 mol H = 0.3416 = 3 =CH3O 1.01g H 0.1156

1.850g O x 1 mol O = 0.1156 = 1 16.00g O 0.1156

Page 17: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Empirical and Molecular Formulas for 500 points

Find the empirical and molecular formula for a compound containing 11.39g Phosphorus and 39.12g

Chloride. Its molar mass is 274.64g.

11.36g P x 1 mol P = 0.3677 = 1 30.97g P 0.3677 39.12g Cl x 1 mol Cl = 1.1035 = 3 PCl3 35.45g Cl 0.3677

P x1 = 30.97 274.64 = 2 = 2(PCl3) = P2Cl6 Cl x3 = +106.35 137.32

137.32

Page 18: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Balancing Equation for 100 points

Balance: _N2 + _H2 _NH3

Balanced: 1N2 + 3H2 2NH3

Page 19: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Balancing Equation for 200 points

Balance: _KClO3 _KCL + _O2

Balanced: 2KClO3 2KCL + 3O2

Page 20: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Balancing Equation for 300 points

Balance: _Na + _H2O _NaOH + _H2

Balanced: 2Na + 2H2O 2NaOH + 1H2

Page 21: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Balancing Equation for 400 points

Balance: _FeCl3 + _NaOH _Fe(OH)3 + _NaCl

Balanced: 1FeCl3 + 3NaOH 1Fe(OH)3 + 3NaCl

Page 22: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Balancing Equation for 500 points

Balance: _C8H18 + _O2 _CO2 + _H2O

Balanced: 2C8H18 + 25O2 16CO2 + 18H2O

Page 23: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Stoichiometry for 100 points

Given this equation:2 KClO3 ---> 2 KCl + 3 O2, How many moles of O2 can be produced by

letting 12.00 moles of KClO3 react?12 mol KClO3 x 3 mol O2 = 18 mol O2

2 mol KClO3

Page 24: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Stoichiometry for 200 points

Given this equation: 2K + Cl2 ---> 2KCl, how many moles of KCl would be produced from

2.50g of K and an excess of Cl2

2.50g K x 1 mol K x 2 mol KCl = 0.0639 mol KCL 39.10g K 2 mol K

Page 25: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Stoichiometry for 300 points

Given this equation: 2NaClO3 2NaCl + 3O2, how many grams of O2 would be produced

from 12.0 moles of NaClO3

12 mol NaClO3 x 3 mol O2 x 32.0g O2 = 576g O2 2 mol NaClO3 1 mol O2

Page 26: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Stoichiometry for 400 points

Given this equation: Na2O + H2O ---> 2NaOH, How many grams of NaOH can be produced

from 54.8 grams of Na2O and an excess of H2O reacting

54.8g Na2O x 1 mol Na2O x 2 mol NaOH x 40 g NaOH = 35.4 61.98g Na2O 2 mol KClO3 1 mol NaOH

Page 27: Stoichiometry Balancing Equations Molecular and Empirical Formulas Percent Composition Mole Conversions 500 400 300 200 100 200 300 400 500 100 200 300.

Stoichiometry for 500 points

Given this equation: 8Fe + S8 ---> 8FeS, How many grams of FeS can be produced from 24.5

grams of S and an excess of Fe reacting24.5g S8 x 1 mol S8 x 8 mol FeS x 87.86 g FeS = 67.2

256.08g S 1 mol S8 1 mol FeS


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