CH 101 Spring 2019
Grams, Moles, and Molecular Mass Review
1. A sample contains 3.4 g of methane (CH4). A. What is the molar mass of methane?
One carbon atom + 4 hydrogen atoms = 12.0108 + 4(1.0079) = 16.0424 g/mol
B. How many moles of methane are in your sample?
3.4 g x (1mol/16.0424g) = 0.21 moles
C. How many molecules of methane are in your sample?
23 23 0.2119 moles x 6.022 x 10 molecules/mole = 1.3 x 10 CH4 molecules
D. How many carbon atoms are in your sample?
23 23 1.3 x 10 CH4 molecules x (1 C atom/molecule) = 1.3 x 10 C atoms
E. How may hydrogen atoms are in your sample?
23 23 1.276 x 10 CH4 molecules x (4 H atoms/molecule) = 5.1 x 10 H atoms
F. What mass of carbon is contained in your sample?
carbon mass/methane mass = (12.0108 g/mol) / (16.0424 g/mol) In 3.4 g methane, there is (12.0108/16.0424) x 3.4g = 2.6 g
Or
0.2119 moles methane contain 0.2119 moles C (because there is 1 C/ CH4) 0.2119 moles C x 12.0108 g/mol C = 2.6 g C
2. A sample contains 6.85 g of carbon dioxide (CO2). A. What is the molar mass of carbon dioxide?
12.0108 g/mol + 2(15.9994 g/mol) = 44.0096 g/mol
B. How many moles of carbon dioxide are in your sample?
6.85 g x (1 mole/44.0096 g) = 0.156 moles
C. How many carbon atoms are in your sample?
0.156 moles CO2 x (1 moles C / 1 mole CO2) = 0.156 moles C 0.156 moles C x 6.022 x 1023 atoms C/mole = 9.37 x 1022 carbon atoms
D. How many oxygen atoms are in your sample?
0.156 moles CO2 x (2 moles O / 1 mole CO2) = 0.312 moles O 0.312 moles O x 6.022 x 1023 O atoms / mole = 1.88 x 1023 O atoms
E. What mass of carbon is contained in your sample?
0.156 moles C x (12.0108 g / mole) = 1.87 g carbon
or
mass of C / mass of CO2 = 12.0108 / 44.0096 in 6.85 g, (12.0108/44.0096) x 6.85 g = 1.87 g carbon
3. A sample contains 5.93 moles of ethanol (CH3CH2OH). A. What is the molar mass of ethanol?
In each ethanol molecule, there are 2 C, 1 O, and 6 H atoms 2 x 12.0108 g/mol + 15.9994 g/mol + 6 x 1.0079 g/mol = 46.0684 g/mol
B. How many grams of ethanol are in your sample?
5.93 moles x (46.0684 g / 1 mole) = 273. g
C. How many molecules of ethanol are in your sample?
5.93 moles x 6.022 x 1023 molecules/mole = 3.57 x 1024 ethanol molecules
D. How many carbon atoms are in your sample?
(2 C atoms/molecule ethanol) x 3.57 x 1024 molecules = 7.14 x 1024 C atoms
E. How many oxygen atoms are in your sample?
(1 O atoms/molecule ethanol) x 3.57 x 1024 molecules = 3.57 x 1024 O atoms
F. How may hydrogen atoms are in your sample?
(6 H atoms/molecule ethanol) x 3.57 x 1024 molecules = 2.14 x 1025 H atoms
G. What mass of carbon is contained in your sample?
(2 moles C/1 mole ethanol) x 5.93 moles ethanol = 11.86 moles C 11.86 moles C x 12.0108 g/mol C = 142. g C
4. A sample contains 10.43 g sodium carbonate (Na2CO3).
A. What is the molar mass of sodium carbonate?
2 Na atoms, 1 carbon atom, and 3 oxygen atoms in each Na carbonate molecule. 2 (22.99 g/mol) + 12.0108 g/mol + 3(15.9994 g/mol) = 105.989 g/mol
B. How many moles of sodium carbonate are in your sample?
10.43 g x (1 mole / 105.989 g) = 0.0984 moles
C. How many moles of carbon are in your sample?
(1 mole C / 1 mole Na bicarbonate) x 0.0984 moles = 0.0984 moles C
D. What mass of carbon is contained in your sample?
0.0984 moles C x 12.0108 g/mol = 1.182 g
or
fraction of C in sample = 12.0108 g/mol / 105.989 g/mol mass of C in sample = (12.0108/105.989) x 10.43 g = 1.182 g
5. A sample contains 82.78 g sodium bicarbonate (NaHCO3).
A. What is the molar mass of sodium bicarbonate?
In each molecule of sodium bicarbonate, there is one sodium atom, one hydrogen atom, one carbon atom, and three oxygen atoms. Overall the molar mass is 22.99 g/mol + 1.0079 g/mol + 12.0108 g/mol + 3(15.9994 g/mol) = = 84.0069 g/mol
B. How many moles of sodium bicarbonate are in your sample?
82.78 g x (1 mole / 84.0069 g) = 0.9854 moles NaHCO3
C. How many moles of oxygen are in your sample?
(3 moles O/mole NaHCO3) x 0.9854 moles NaHCO3 = 2.956 moles oxygen atoms
D. How many moles of carbon are in your sample?
(1 mole C/mole NaHCO3) x 0.9854 moles NaHCO3 = 0.9854 moles C
E. What mass of carbon is contained in your sample?
0.9854 moles C x 12.0108 g/mole = 11.84 g C
or
fraction of C/NaHCO3 = 12.0108 g/mol / 84.0069 g/mol amount of C in sample = (12.0108/84.0069) x 82.78 g = 11.84 g