Molar Mass & Percent Composition
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Transcript Molar Mass & Percent Composition
September 20 2011
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
To calculate molar mass…
◦ Add the atomic masses of each atom in a
compound and switch the label from AMU to gram
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
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.
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%
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%
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%
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
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
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