Question #1
Empirical & Molecular
W/S
Solutions (#1, 4, 5, 8-10)
Step 1 (calculate M for the empirical formula)
C2H5= ! 2 x 12.01g/mol
The empirical formula of butane, the
fuel used in disposable lighters, is
C2H5. In an experiment, the molar
mass of butane was determined to be
58g/mol. What is the molecular
formula of butane?
Step 2 (divide M by the mass of the empirical
formula
! ! ! 5 x 1.01g/mol
58g/mol / 29.07g/mol = 2.0
! ! ! 29.07g/mol
Therefore, if your empirical formula is
Ask yourself: “how many times will the mass of my
empirical formula fit into the molar mass as
mass ______________?”
spectrometer
determined by the _____
Then your molecular formula is
~
2x
C2H5
C2H5 x 2 =
C4H10
Butane!!!
Question #4
Step 1 (use % composition to obtain mass)
A compound’s molar mass is
240.28g/mol. Its percentage
composition is 75.0% carbon, 5.05%
hydrogen, and 20.0% oxygen.
What is the compound’s molecular
formula?
Calculate the number of moles of each
represented element:
nC: 1.80 x 102g / 12.01g/mol = 15.0 mol
~
12
nO: 48.0g / 16.00g/mol = 3.00 mol
C15H12O3
nC: 0.750 (75.0%) x 240.28g = 180g of C
(1.80 x 102g)
nH: 0.0505 (5.05%) x 240.28g = 12.1g of H
nO: 0.200 (20.0%) x 240.28g = 48.0g of O
Question #5
Step 2 (mass to moles - m to n)
nH: 12.1g / 1.01g/mol = 12.0 mol
Calculate the number of grams of each
represented element using % composition:
~
3
~
15
Tartaric acid, also known as cream
of tartar is used in baking. Its
empirical formula is C2H3O3. If
1.00 mol of tartaric acid contains
3.61 x 1024 oxygen atoms, what is
the molecular formula of tartaric?
Step 1 (find # of moles of oxygen)
Ntartaric: 3.61 x 1024 atoms x 1 mol/6.02 x 1023
atoms = 6.00 mol of oxygen in one mole of
tartaric acid
Step 2 (relationship with empirical formula)
Question #8
What is the percent by mass of water
in magnesium sulfite hexahydrate?
If you have 6 moles of oxygen, then the molecular
formula is 2 x C2H3O3 =
C4H6O6
Divide the mass of water by the mass of the
entire compound:
6H2O
MgSO3 6H2O
108.12g/mol / 212.49g/mol = 0.50882
= 50.882 %
Question #9
A 3.34g sample of a hydrate has the
formula SrS2O3 xH2O, and
contains 2.30g of SrS2O3. Find the
value of x.
Step 1 (find the # of moles of SrS2O3)
nSrS2O3: 2.30g / 199.74g/mol = 0.0115 mol SrS2O3
Step 2 (find the # of moles of H2O)
If the total sample is 3.34g and 2.30g is SrS2O3,
then 3.34g - 2.30g = 1.04g of H2O
nH2O: 1.04g / 18.02g/mol = 0.0577 mol H2O
Step 3 (solve for X)
Find the relationship between SrS2O3 and H2O:
1 mol(SrS2O3) = X mol(H2O)
and
0.0115 mol(SrS2O3) = 0.0577 mol(H2O)
0.0577 mol
Therefore: X mol
=
1 mol
0.0115 mol
X = 5.02
~
5 mol of H2O
SrS2O3
Question #10
A hydrate of zinc chlorate,
Zn(ClO3)2 XH2O, contains 21.5%
zinc by mass. Find the value of X.
5H2O
Step 1 (assume 100g sample, find the # of moles
of Zn)
nZn: 21.5g / 65.38g/mol = 0.329 mol Zn
Step 2 (use moles of Zn to find moles of ClO3)
In Zn(ClO3)2 there are 2 moles of ClO3 for every
1 mole of Zn
nClO3: 0.329 mol(Zn) x 2(ClO3)/1(Zn) = 0.658 mol ClO3
Step 3 (use moles of ClO3 to calculate mass)
mClO3: 0.658 mol x 83.45g/mol = 54.9g of ClO3
Step 4 (find the mass of water)
In a 100g sample, we have 21.5g(Zn) and 54.9g(ClO3),
so: mH2O: 100g - 21.5g - 54.9g = 23.6g of H2O
Step 6 (solve for X)
Find the relationship between Zn and H2O:
1 mol(Zn) = X mol(H2O)
and
0.329 mol(Zn) = 1.31 mol(H2O)
Step 5 (find moles of H2O)
Therefore: X mol
1.31 mol
=
1 mol
0.329 mol
nH2O: 23.6g / 18.02g/mol = 1.31 moles of H2O
X = 3.98
~
4 mol of H2O
Zn(ClO3)2 4H2O