Download 4 - Emp. and Mol. Form -q

SCH3U
Unit: Quantities in Chemical Reactions
Empirical Formula and Molecular Formula (Nelson, pg. 185-193)
Name: ___________
Date: ___________
The Empirical Formula of a compound is the simplest formula which can represent the ratio of atoms present.
The Molecular Formula of a compound is a list of all the atoms which actually make up a compound. The numbers of
atoms present follow the ratio predicted by the empirical formula.
NOTE: Sometimes the molecular formula is the same as the empirical formula. Sometimes one empirical formula
can have more than one possible molecular formula.
Examples:
Empirical Formula
Molecular Formula
NH2
CH
N2H4
C6H6
CH2O
H2O
Sample Problem 1:
C6H12O6
H2O
Finding an Empirical Formula from Percentage Composition Data
A compound is 50.91% Zn, 16.04% P and 33.15% O. What is the empirical formula of the compound?
Solution:
Because only the relative amounts of the elements are needed for an empirical forumula, any absolute amountof the
compound can be used. So, to make things “easy”, assume that you have a mass which is a nice round number like
100 g. In 100 g of sample, there would be 50.91 g Zn, 16.04 g P and 33.15 g O.
n Zn
= mass of Zn
molar mass of Zn
=
50.91 g
65.41 g/mole
= 0.7783 moles
nP
= mass of P
molar mass of P
=
16.04 g
30.97 g/mole
= 0.5179 moles
nO
=
mass of O
molar mass of O
=
33.15 g
16.00 g/mole
= 2.0718 moles
Initial Empirical Formula = Zn0.7783P0.5179O2.0718
Molecular formulas and formula units are always given in whole number values. Divide through by the smallest
number of moles in the ratio (in this case that’s 0.5179 mol) to find the whole number ratio of atoms in the
empirical formula. Note that showing the mole unit here is optional as all of the units will cancel out during the
division.
Zn 0.7783 P0.5179 O 2.0718 = Zn1.5P1O4
0.5179
0.5179
0.5179
In many calculations, the subscripts will be all whole numbers by the time this step is over. In this case, the
subscript for Zn is not a whole number (it is 1.5) so we will need to multiply all of the subscripts by the same whole
number value to make 1.5 into the smallest whole number possible. The multiplier is the smallest whole number that
will accomplish the task – it this case, it is 2.
Zn 1.52 P1x2O4x2  Zn3PO
2 8
There are 3 different elements in this compound, one metal and 2 non – metals so this is likely an ionic compound
containing a phosphate ion. The most likely empirical formula then is…Zn3(PO4)2.
Note that the problem above shows all the possible challenges you may run into when solving. Not all problems will
show all of these features.
NOTE: A summary of the steps involved in determining empirical formula is found at the
bottom of page 188.
1.
2.
Do #1-5 at the bottom of page 186. Do #4 page 197
Read over the sample problem in the note above and Sample Problem 1 page 188. Make your own notes from
the text Sample Problem as is necessary. Do #3-6 at the top of page 189.
SCH3U
Unit: Quantities in Chemical Reactions
Empirical Formula and Molecular Formula
Name: ___________
Date: ___________
Sample Problem 2:
Finding a Molecular Formula from the Empirical Formula and the Molar Mass of
the Empirical Formula
𝑔
CH2 is the empirical formula of a certain compound whose molar mass is 42.0
. What is the molecular formula of this
𝑚𝑜𝑙
compound?
Solution:
Note: The mass of a molecular formula may be the same as the mass of its empirical formula. This happens when
the two chemical formulas are identical. If the formulas are different from one another, the molecular formula will
be a multiple of the empirical formula so the moleculear formula mass will be a multiple of the empirical formula
mass.
Mass of empirical formula, CH2
= 12.0 g of C + (2x 1.0 g of H)
= 14.0
𝑔
𝑚𝑜𝑙
Mass of molecular formula = Empirical Mass x Whole Number which represents the ratio of molecular mass to empirical
mass
Whole Number = Molecular Mass
Empirical Mass
𝑔
42.0
𝑚𝑜𝑙
=
𝑔
14.0
𝑚𝑜𝑙
= 3
This means the mass of the
molecular formula is 3 times larger
than the the mass of
the empirical formula.
Because the molecular mass is 3X the value of the empirical mass, the molecular formula must be 3 times the
empirical formula of CH2.
So, the molecular formula is CH2 X 3 = C3H6.
Practice:
1. The empirical formula for glucose is CH2O(s). The molar mass of glucose is 180.18 g/mol. Determine the
molecular formula for glucose.
(Answer: C6H12O6)
2. The empirical formula for xylene is C4H5(l) and its molar mass is 106 g/mol. Determine the molecular formula
for xylene.
(Answer: C8H10(l))
3. The empirical formula for the compound 1, 4-butanediol is C2OH2(l). The molar mass of the compound is
90.14 g/mol. What is the molecular formula for the compound?
(Answer: C4O2H10(l))
4. The empirical formula for styrene is CH(l). The molar mass of the molecule is 104 g/mol. Determine the
molecular formula ? (Answer: C8H8(s))
Sample Problem 3:
Finding a Molecular Formula from Mass Percent (Percentage Compostion) Data
Analysis shows that the butane fuel in a butane lighter is 82.5% carbon and 17.5% hydrogen by mass. The molar
mass of butane is found to be 58.0 g. What is the molecular formula of butane?
Solution:
We need to determine the number of moles of carbon and hydrogen in the sample of butane. If we assume we have
82.5
17.5
100 g of butane, then 82.5% of the sample or
parts of it will be carbon by mass and 17.5% or
will be
100
100
hydrogen by mass. Mathematically,
Carbon
Hydrogen
82.5
mC =
x 58.0 g sample
100
= 47.8 g of carbon
17.5
mH =
x 58.0 g sample
100
= 10.2 g
m
nC =
mm
m
nH =
mm
=
47.8 g
g
12.0
mol
= 3.98 mol of carbon
=
10.2 g
g
1.01
mol
= 10.0 mol of hydrogen
In this instance, the ratio of atoms is 4 C : 10 H so the empirical formula is C4H10 .
Practice
NOTE:
Page 193 presents a good summary of the types of molecular formula problems.
Do #8, 9, 10 page 193. Try #6 page 197.
A Similar Type of Problem: Determining the Formula for A Hydrate
Some Definitions
We learned how to name and write formulas for hydrdates in Unit 1 of SCH 3U.
Term
Hydrate
Anhydrous form of a hydrate
Definition and Example
A hydrate is an ionic compound that has a specific number of water molecules
bound to each formula unit. Often, when a crystal forms from a water solution,
water molecules aer trapped in the crystal in a specific arrangement.
Eg CaSO4•2H2O(s) is the formua for gypsum crystals. For each formula unit of
calcium sulfate there are thwo water molecules.
Note that the water molecules will add mass to the solid.
An anydrous form of a hydrate is the ionic compound without the trapped water.
So the anydrous form of the hydrate CaSO4•2H2O(s) is CaSO4(s). Note that
usually, the anydrous form of the solid can be made by heating the hydrate to
drive away the water.
Sample Problem 4
Determine the percent by mass of water in LiCl•4H2O(s).
Solution
mmhydrate = mmLiCl + 4(mmH2O )
= [(6.94) + (35.45) + 4(1.02 + 16.00)]
= 114.39
g
So, water makes up about 59.5% of
the hydrate by mass.
g
mol
mol
% 𝑏𝑦 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻2 𝑂 𝑖𝑛 ℎ𝑦𝑑𝑟𝑎𝑡𝑒 =
=
𝑚𝑎𝑠𝑠 𝑤𝑎𝑡𝑒𝑟 𝑖𝑛 ℎ𝑦𝑑𝑟𝑎𝑡𝑒
× 100
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 ℎ𝑦𝑑𝑟𝑎𝑡𝑒
4(1.02+16.00)𝑔
114.39 𝑔
× 100
= 59.5156%
Sample Problem 5
Determine the formula for the hydrate of chromium(III) nitrate that is 40.50% water by mass.
Solution
The formula for chromium(III) nitrate is Cr(NO3)3 so its hydrate is Cr(NO3)3•xH2O where x represents the
number of moles of water associated with one mole of chromium(III) nitrate.
mmchromium ( III ) nitrate  (52.00)  3(14.01)  9(16.00)
 238.03
g
mol
m
mm
59.50g

g
238.03
mol
=0.2499mol
nchromium( II ) nitrate 
nwater 

m
mm
40.50 g
[2(1.01)+16.00]
= 2.2475 mol
g
mol
The hydrate is 40.50% water by mass.
So, if you assume you have exactly 100 g of the hydrate, 40.50 g
will be water and the rest, 59.50 g, will be the Cr(NO3)3 part of
the hydrate. On the left side of this box, the calculations are
shown for determining how many moles of Cr(NO3)3 and H2O are
present in the hydrate based on these mass values.
Based on the number of moles of Cr(NO3)3 and H2O present in
the hydrate, the formula for the hydrate would be
0.2499 Cr(NO3)3•2.2475 H2O but we need the mole values to be
in whole numbers and we need to be showing 1 mole of Cr(NO3)3,
so we divide through by the number of moles of Cr(NO3)3…
0.2499
2.2475
Cr  NO3 3 •
H 2O  Cr(NO3)3 •9H2O
0.2499
0.2499
So, the formula for the hydrate is Cr(NO3)3•9H2O.
Practice
1. For each of the following hydrates, calculate the percent by mass of water.
a) MgSO3•6H2O(s) (Ans: 50.88%)
b) CaSO4•2H2O(s) (Ans: 13.43%)
2. A 3.34 g sample of a hydrate, SrS2O3 • xH2O(s), contains 2.30 g of SrS2O3(s). Find the value of x. (Ans: 5)
3. A hydrate of zinc chlorate, Zn(ClO3)2 • xH2O(s), contains 21.3% zinc by mass. Find the value of x. (Ans: 4)