Download Chapter 3: Calculations with Chemical Formulas

CHAPTER 3: CALCULATIONS WITH
CHEMICAL FORMULAS
AND EQUATIONS
Vanessa Prasad-Permaul
Valencia College
CHM 1045
Atomic and Molecular Mass
 The molecular mass is the sum of the masses
of the atoms making up the molecule (units
amu) (molecular compounds only!)
calculate the molecular mass:
H2O
C2H4O2
Formula Mass
 Formula Mass is the sum of atomic masses of all
atoms in a formula unit of any compound,
molecular or ionic (units amu)
calculate the formula mass:
NaCl
MgCO3
Exercise 3.1
 Calculate the formula masses of the following
compounds, using the table of atomic masses. Give
the answers to 3 significant figures:




A) NO2
B) glucose, C6H12O6
C) sodium hydroxide
D) magnesium hydroxide
A. NO2
1 x AM of N
2 x AM of O
MM of NO2
=
=
=
2 x 15.9994
B. C6H12O6
6 x AM of C
12 x AM of H
6 x AM of O
MM of C6H12O6
=
=
=
6 x 12.011=
12 x 1.0079
6 x 15.9994
=
C. NaOH
1 x AM of Na
1 x AM of O
1 x AM of H
MM of NaOH
14.0067
=
31.9988
46.0055
46.0 amu
72.066
=
=
180.1572
180. amu
=
=
=
=
12.0948
95.9964
22.98977
15.9994
1.0079
39.9971
40.0 amu
D. Mg(OH)2
1 x AM of Mg
2 x AM of O
2 x AM of H
MM of Mg(OH)2
=
=
=
=
2 x 15.9994
2 x 1.0079
=
= 2.0158
58.3196
58.3 amu
24.305
31.9988
Avogadro
 One mole of a substance is the quantity of a
given substance that contains as many
molecules or formula units as the number of
atoms in exactly 12g of C-12.
 One mole of carbon = 6.02 x 1023 molecules of
carbon
 Avogadro’s Number (NA = 6.022 x 1023
molecules) is the numerical value assigned to
the unit, 1 mole
Avogadro’s Number
Li2SO4
1. How many Li2SO4 units are in 3 moles of Li2SO4?
2. How many Li atoms are in 3 moles of Li2SO4?
3. How many S atoms are in 3 moles of Li2SO4?
4. How many O atoms are in 3 moles of Li2SO4?
Molar Mass of a substance is the mass of one mole of
that substance
C-12 has a molar mass of exactly 12g/mol
For all substances, the molar mass (g/mol) is numerically
equal to the formula mass (amu).
For example:
1 mole O contains 6.02 x 1023 oxygen atoms
1 mole of O2 contains 2x (6.02 x 1023) oxygen atoms
Molar Mass
Exercise 3.3
What is the mass in grams of a calcium atom?
What is the mass in grams of an ethanol molecule?
Molar Mass
=
The atomic mass of Ca = 40.08 amu; thus, the molar mass = 40.08
g/mol, and 1 mol Ca = 6.022 x 1023 Ca atoms.
Mass of one Ca =
40.08 g
1 mol Ca
x
1 mol
6.022 x 1023 atoms
= 6.6556 x 10−23
= 6.656 x 10−23 g/atom
The molecular mass of C2H5OH, or C2H6O, = (2 x 12.01) + (6 x
1.008) + 16.00 = 46.068. Its molar mass = 46.07 g/mol, and 1 mol
= 6.022 x 1023 molecules of C2H6O.
Mass of one C2H6O =
46.07 g
1 mol C2 H6 O
1 mol
x 6.022 x 1023 molecules
= 7.6503 x 10−23
= 7.650 x 10−23 g/molecule
Mole Calculations
Consider ethanol with a molar mass of 46.1g/mol.
Conversion to moles of ethanol:
1 mol C2H5OH = 46.1g/mol
Prepare acetic acid from 10.0g of ethanol. How many
moles of ethanol is this?
10.0g C2H5OH x 1 mol C2H5OH = 0.217 mol C2H5OH
46.1 g C2H5OH
Mole Calculations
Exercise 3.4
Dilute aqueous hydrogen peroxide is a colorless
liquid used as a bleach. Analysis of a solution shows it contains
0.909 mol of H2O2 in 1.00L of solution. What is the mass of
hydrogen peroxide in this volume of solution?
The molar mass of H2O2 is 34.02 g/mol. Therefore,
34.02 g H2 O2
0.909 mol H2 O2 x
= 30.92 = 30.9 g H2 O2
1 mol H2 O2
Mole Calculations
Exercise 3.5
HNO3 is a corrosive liquid used for the manufacture of
nitrogen fertilizers and explosives. A sample containing 28.5g
of nitric acid was poured into a beaker. How many moles of
HNO3 are there?
The molar mass of HNO3 is 63.01 g/mol. Therefore,
1 mol HNO3
28.5 g HNO3 x
= 0.4523 = 0.452 mol HNO3
63.01 g HNO3
Mole Calculations
Exercise 3.6
HCN is a volatile colorless liquid. The compound is highly
poisonous . How many molecules are there in 56mg HCN,
the average toxic does?
Convert the mass of HCN from milligrams to grams.
Then convert grams of HCN to moles of HCN.
Finally, convert moles of HCN to the number of HCN molecules.
56 mg HCN x
6.022 x 1023 HCN molecules
1 mol HCN
1g
x 27.02 g HCN x
1000 mg
1 mol HCN
= 1.248 x 1021 = 1.2 x 1021 HCN molecules
Chemical Formulas
• Percent Composition: Identifies the elements
present in a compound as a mass percent of the
total compound mass.
Mass % A = mass A in the whole X 100%
whole mass
Exercise 3.7
Ammonium nitrate, which is prepared from nitric acid is
used as a nitrogen fertilizer. Calculate the mass percentages
of the elements in ammonium nitrate to 3 sig. figs.
The molecular mass of NH4NO3 = 80.05; its molar mass = 80.05g/mol.
Hence
Percent N =
28.02g N
x 100% = 35.00 = 35.0%
80.05g NH4NO3
Percent H =
100%
= 5.036 = 5.04%
4 .032g H
x
80.05g NH4NO3
Percent O =
48.00g O
80.05g NH4NO3
x
100% = 59.96 = 60.0%
Exercise 3.8
How many grams of nitrogen are there in a fertilizer
containing 48.5g of NH4NO3?
From the previous exercise, NH4NO3 is 35.0% N
(Fraction N = 0.350), so the mass of N in 48.5 g of NH4NO3 is:
48.5 g NH4NO3 x
0.350 g N
1 g NH4NO3
= 16.975 = 17.0 g N
Exercise 3.9
A 3.87 mg of ascorbic acid (vitamin C) gives 5.80mg of CO2 and 1.58mg of
H2O on combustion. What is the % composition of this compound?
First convert the mass of CO2 to moles of CO2. Next, convert this to moles of C
(1 mol CO2 is equivalent to 1 mol C). Finally, convert to mass of C,
changing milligrams to grams first:
5.80 x 10−3 g CO2 x 1 mol CO2 x
44.01g CO2
1.583mg C
1 mol C x 12.01g C = 1.583 x 10−3g C
1 mol CO2
1 mol C
x 100% = 40.9% C
3.87mg
1.58 x 10-3g H2O x 1 mol H2O x 2 mol H
x 1.0079g H = 1.767 x 10-4g H
18.02g H2O 1 mol H2O
1 mol H
1.767 x 10-4g H x 100% = 4.57% H
3.87 x 10 -3g
Now that we have the % composition of C & H, we can
calculate % O:
1.583mg C
x
100% = 40.9% C
3.87mg
1.767 x 10-4g H x 100% =
4.57% H
3.87 x 10 -3g
The mass percentage of O can be determined by subtracting the sum of
the above percentages from 100%:
Percent O = 100.000% − (40.90 + 4.5658) = 54.5342 = 54.5% O
Chemical Formulas
• Empirical Formula: Determined from data
about percent composition, tells only the
smallest whole number ratio of atoms in a
compound.
• Tells the ratio of numbers of atoms in the
compound
• hydrogen peroxide
• Molecular formula is H2O2
• Empirical formula is HO
Chemical Formulas
Exercise 3.10
A sample of compound weighing 83.5g contains 33.4g of sulfur.
The rest is oxygen. What is the empirical formula?
Convert the masses to moles that are proportional to the subscripts
in the empirical formula:
83.5g compound – 33.4g S = 50.1g O
33.4g S x 1 mol S = 1.0414 mol S
32.07g S
50.1g O x 1 mol O = 3.1312 mol O
15.9994g O
Next, obtain the smallest integers from the moles by dividing each by the smallest
number of moles:
For O: 3.1312 = 3.01 (3 oxygen)
1.0414
For S: 1.0414 = 1.00 (1 sulfur)
1.0414
The empirical formula is SO3
Chemical Formulas
Exercise 3.11
Benzoic acid is a white crystalline powder use a food preservative.
The compound contains 68.8% C, 5.0% H and 26.2% O by mass.
What is the empirical formula?
For a 100.0-g sample of benzoic acid, 68.8 g are C, 5.0 g are H, and 26.2
g are O. Using the molar masses, convert these masses to moles:
68.8g C x 1 mol C = 5.729 mol C
12.01g C
5.0g H x 1 mol H = 4.961 mol H
1.0079g H
26.2g O x 1 mol O = 1.638 mol O
15.9994 g O
5.759 = 3.49 C
1.638
4.961 = 3.03 H
1.638
1.638 = 1.00 O
1.638
Rounding off, we obtain C3.5H3.0O1.0. Multiplying the numbers by 2 gives whole numbers,
for an empirical formula of C7H6O2.
• Molecular Formula: a multiple of the empirical formula
which is obtained by summing the atomic masses of the
atoms in the empirical formula.
Molecular mass = n x empirical formula mass
n = the number of empirical formula units in the molecule. The
molecular formula is obtained by multiplying the subscripts of
the empirical formula by n.
Exercise 3.12
The % composition of acetaldehyde is 54.5% C, 9.2% H and 36.3%
O. The molecular mass is 44 amu. What is the molecular formula of
acetaldehyde?
For a 100.0-g sample of acetaldehyde, 54.5 g are C, 9.2 g are H, and 36.3 g
are O. Using the molar masses, convert these masses to moles:
54.5g C x 1 mol C
12.01g C
= 4.537 mol C
4.537 = 2.0 C
2.270
9.2g H x 1 mol H = 9.127 mol H
1.0079g H
9.127 = 4.0 H
2.270
36.3g O x 1 mol O = 2.270 mol O
15.994 g O
2.270 = 1.0 O
2.270
The empirical formula of acetaldehyde is C2H4O
The empirical formula of acetaldehyde is C2H4O.
The empirical formula mass is 2C x 12 amu = 24 amu
4H x 1 amu = 4 amu
1O x 16 amu = 16 amu
44 amu
To find n = molecular mass
= 44 amu = 1
empirical formula mass 44 amu
Therefore the molecular formula is the same as the empirical
formula
Stoichiometry
• Stoichiometry: Relates the moles of products and
reactants to each other and to measurable quantities
involved in a chemical reaction.
Convert
moles of A to
moles of B
Grams of A
Convert
grams of A to
moles of A
Grams of B
Convert
moles of B to
grams of B
Stoichiometry
Exercise 3.14
Sodium is a soft reactive metal that instantly reacts with
water to give hydrogen gas and a solution of sodium hydroxide.
How many grams of sodium metal is needed to give 7.81g of
hydrogen by this reaction:
Equation: 2Na + 2H2O  H2 + 2NaOH
From this equation, 2 moles of Na corresponds to one 1 mole of H2.
Therefore,
7.81 g H2 x 1 mol H2 x 2 mols Na x 22.9898g Na = 178.1 = 178 g Na
2.0158g H2 1 mol H2
1 mol Na
Stoichiometry
Exercise 3.15
Zinc sulfide reacts with oxygen to give zinc oxide and sulfur
Dioxide. How many kg of oxygen gas combines with
5.00 x103 g of zinc sulfide in this reaction?
2 ZnS + 3 O2
2 ZnO + 2 SO2
5.00 x 103g ZnS x 1 mol ZnS x 3 mols O2 x 31.999g O2
97.474g ZnS 2 mols ZnS 1 mol O2
= 2462g O2 x 1kg
1000g
= 2.46kg O2
Stoichiometry
• Limiting Reagent: The reactant or the reagent that is
completely consumed in a chemical reaction
– I need to make a fruit salad that is 1/2 apples and
1/2 oranges. I have 10 apples but only 7 oranges.
What is the limiting fruit? How many apples and
oranges can I use?
Exercise 3.17
Aluminum chloride is used as a catalyst in various industrial
reactions. It is prepared from hydrogen chloride gas and aluminum metal
shavings. A reaction vessel contains 0.15 mol Al and 0.35 mol HCl. How
many moles of AlCl3 can be prepared from this mixture?
2 Al (s) + 6 HCl (g)
2 AlCl3(s) + 3 H2(g)
First, determine the limiting reactant by calculating the moles of AlCl3 that would be
obtained if Al and HCl were totally consumed:
0.15 mol Al x 2 mol AlCl3 = 0.15 mol AlCl3
2 mol Al
0.35 mol HCl x 2 mol AlCl3 = 0.12 mol AlCl3 (HCl is the limiting reactant)
6 mol HCl
Because the HCl produces the smaller amount of AlCl3, the reaction will stop when HCl is totally
consumed but before all the Al is consumed. The limiting reactant is therefore HCl. The amount
of AlCl3 produced must be 0.1166, or 0.12 mol.
Stoichiometry
• Limiting Reagent Calculation: Lithium oxide is a
drying agent used on the space shuttle. If 80.0 kg
of water is to be removed and 65 kg of lithium
oxide is available, which reactant is limiting?
Li2O(s) + H2O(l)  2 LiOH(s)
Li2OMM = 29.88 g/mol
H2OMM = 18.02 g/mol
Stoichiometry
• Yields of Chemical Reactions: If the actual amount of
product formed in a reaction is less than the theoretical
amount, we can calculate a percentage yield.
• Theoretical Yield: maximum amount of product that can be
obtained by a reaction from the given amounts of reactants.
• Percent Yield: the actual yield (experimentally determined)
expressed as a percentage of the theoretical yield
(calculated).
% Yield = Actual yield
Theoretical yield
X 100%
Stoichiometry
Exercise 3.19
New industrial plants for acetic acid react liquid
methanol with carbon monoxide in the presence
of a catalyst.
CH3OH(l) + CO (g)
HC2H3O2(l)
15.0g of methanol and 10.0g of carbon
monoxide were placed in a reaction vessel.
What is the theoretical yield of acetic acid. If the
actual yield is 19.1g, what is the percent yield?
Stoichiometry
Exercise 3.19
Convert grams of each reactant to moles of acetic acid:
15.0g CH3OH x 1 mol CH3OH x 1 mol HC2H3O2 = 0.468mol HC2H3O2
32.04g CH3OH 1 mol CH3OH
10.0g CO x 1 mol CO x 1 mol HC2H3O2 = 0.357mol HC2H3O2 (limiting)
28.01g CO
1 mol CO
Thus, CO is the limiting reactant, and 0.0357 mol HC2H3O2 is obtained. The
mass of product is
0.357mol HC2H3O2 x 60.05g HC2H3O2 = 21.44g HC2H3O2 (theoretical)
1 mol HC2H3O2
The percentage yield is
19.1g x 100% = 89.08 = 89.1%
21.44g
Stoichiometry
• CH2Cl2 is prepared by reaction of CH4 with Cl2 giving
HCl. How many grams of CH2Cl2 result from the
reaction of 1.85 kg of CH4 if the yield is 43.1%?
CH4(g) + 2 Cl2(g)  CH2Cl2(l) + 2 HCl(g)
1.85kg CH4 x 1000 g x 1 mol CH4 x 1 mol CH2Cl2 = 115.62 mol CH2Cl2
1 kg
16 g
1 mol CH4
115.62 mol CH2Cl2 x 84 g = 9712.5 g or 9.71kg CH2Cl2 produced
1 mol
43.1% =
Stoichiometry
K2PtCl4 + 2 NH3  Pt(NH3)2Cl2 + 2 KCl
• If 10.0 g of K2PtCl4 and 10.0 g of NH3 are allowed
to react:
a) which is limiting reagent?
b) how many grams of the excess reagent are
consumed?
c) how many grams of cisplatin are formed?
K2PtCl4 = 415.08 g/mol NH3 = 17.04 g/mol
Pt(NH3)2Cl2 = 300.06 g/mol
Percent Composition
What is glucoses, C6H12O6, empirical formula, and
what is the percentage composition of glucose?
Percent Composition
• Saccharin has the molecular formula C7H5NO3S, what is its
empirical formula and the percent composition?
Empirical Formula
• A compound was analyzed to be 82.67% carbon
and 17.33% hydrogen by mass. The molar mass is
58.11 g/mol.
– What is the empirical formula and molecular
formula?