Molar Mass & Percent Composition

Report
September 20 2011
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Molar mass – the mass of one mole of a
substance
◦ Gram formula mass (gfm) – mass in grams of one
mole of an ionic compound
◦ Gram molecular mass (gmm) – mass in grams of
one mole of a molecular compound
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To calculate molar mass…
◦ Add the atomic masses of each atom in a
compound and switch the label from AMU to gram
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Find the gram formula mass of beryllium oxide
◦ Beryllium oxide = BeO
Be = 9.0 AMU
O = 16.0 AMU
Total = 25.0 AMU -> 25.0 grams
◦ Lithium sulfide = Li2S
◦ Li = 4.0 AMU x 2 = 8.0 AMU
◦ S = 32.1 AMU
 8.0 + 32.1 = 40.1 AMU -> 40.1 g
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If you have 100 coins and 30 of them are
quarters, what percent of all your coins is
quarters?
percent = part/whole x 100
part = 30 and whole = 100
30/100 x 100= 30%
Percent Composition: the percent by mass of each element in a compound
To calculate percent composition, find the mass of the element and
divide that by the total mass of the compound.
% composition = mass of element x100
mass of compound
In a sample of hydrogen peroxide, the mass of hydrogen is 2.0 grams
and the total mass of the compound is 34.0 grams. Find the percent
composition of hydrogen in hydrogen peroxide.
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8.20 g of magnesium combines with 5.40 g
of oxygen to form a compound.
Find the percent composition of this compound…
Total mass = 8.20g + 5.40g = 13.6g
%Mg = 8.20g/13.6g x100 = 60.3%
%O = 5.40g/13.6g x100 = 39.7%
◦ Check… 60.3% + 39.7% = 100%
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29.0 g Ag combine with 4.30 g S to form a
compound. Find the percent composition.
◦ Hint… don’t forget to find the total mass of the whole compound!
◦ Total mass = 29.0g + 4.30g = 33.30g
◦ %Ag = 29.0g/33.30g x100 = 87.1%
◦ %S = 4.30g/33.30g x100 = 12.9%
 Check…. 87.1% + 12.9% = 100%
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Propane is C3H8. Use molar mass to find
percent composition.
◦ Use the periodic table to find the atomic mass of each element…
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C=12.0AMU and H = 1.0AMU and change that to molar mass
C = 12.0g and H = 1.0g
Three carbon atoms in the formula = 12.0g x 3 = 36.0g
Eight hydrogen atoms = 1.0g x 8 = 8.0g
Total molar mass of compound = 36.0 + 8.0 = 44.0g
%C = 36.0g/44.0g x100 = 81.8%
%H = 8.0g/44.0g x100 = 18.2%
◦ Check… 81.8% + 18.2% = 100%
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What is the mass of carbon in 82.0 g propane?
◦ Hint… use carbon’s percent composition!
◦ We just found that the percent of carbon in propane is
81.8%. That means that 81.8% of the mass of a sample of
propane will be carbon.
◦ So to calculate the mass of carbon in 82.0 g, just multiply
by the percent, 81.8%. Don’t forget to change the percent
to a decimal!
◦ 82.0g x 0.818 = 67.1 g carbon
 That means that in 82.0g propane, there are 67.1 g carbon
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Empirical Formula:
formula with the lowest wholenumber ratio of elements in a compound
◦ In other words… reduce the fraction!
◦ Glucose : C6H12O6
◦ Benzene: C6H6
◦ Ribose: C5H10O5
ratio is 6:12:6… you can divide each of those by 6 and get 1:2:1
so the empirical formula is C1H2O1 or CH2O
ratio is 6:6, divide both by 6 and get 1:1…. CH
ratio is 5:10:5, divide each by 5 and get 1:2:1…. CH2O
S2Cl2
divide both by 2… SCl
C6H10O4
divide all by 2, C3H5O2
H2O2
HO
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Find percent composition
◦ Carbon is black, oxygen is red, hydrogen is white
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Use the periodic table to mind atomic mass…. C=12.0AMU, O=16.0AMU, H=1.0AMU
Count the atoms to write the chemical formula… C3H6O
Molar mass: C=12.0g x 3 = 36.0g
H=1.0g x 6 = 6.0g
O = 16.0g
Add all the masses to find total gram molecular mass… 36.0g+6.0g+16.0g=58.0g
%C=36.0g/58.0g = 62.1%
%H=6.0g/58.0g = 10.3%
%O=16.0g/58.0g = 27.6%
Check… 62.1%+10.3%+27.6%=100% 
Write the empirical formula of acetone… C3H6O… can’t be reduced

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