Download Chapter 3: Stoichiometry

Chapter 3: Stoichiometry
I.
3.1: Atomic Masses
The modern system of atomic masses is carbon 12. Carbon 12 has exactly 12 atomic mass units. The most
accurate method for comparing atomic mass is through the use of the mass spectrometer. The mass of
Carbon 13 divided by the mass of Carbon 12 equals 1.0836129. Therefore the mass of Carbon 13 =
(1.0836129)(12 amu) = 12.003355 amu. The average atomic mass is called the atomic weight.
3.2: The Mole
The mole is used to count atoms. It is the number equal to the number of carbon atoms in exactly 12 grams
of pure Carbon 12. The numerical value of a mole is 6.022 x 10^23 units. This number is called
Avogadro’s number. Avogadro’s Constant uses the unit of atoms. So if (6.022 x 10^23 atoms)(12
amu/atom)=12 grams, then 6.022 x 10^23 amu = 1 gram.
3.3: Molar Mass
The molar mass of a substance is the mass in grams of one mole of the compound. The molar mass of a
known substance is obtained by summing the masses of the component atoms.
3.4: Percent Composition of Compounds
There are two ways to describe the composition of a compound. We can obtain the mass percent of the
elements from the formula of the compound by comparing the mass of each element present in 1 mole of
the compound to the total mass of 1 mole of the compound.
3.5: Determing the Formula of a Compound
It is determined by taking a weighed sample of the compound and either decomposing it into its component
elements or reacting it with oxygen to produce substances such as CO2, H2O, and N2 which are then
collected and weighed. Once you find the moles of every element, divide all by the lowest mole number.
Multiply all three to gain whole numbers. The end result will be the empirical formula. To find the
molecular formula, we must know the molar mass. The molar mass divided by the empirical formula mass
gives a whole number. Multiply the number to the empirical formula to find the molecular formula.
3.6: Chemical Equations
In a chemical equation, the right side is called the products and the left is called the reactants. To be sure
that the elements on the reactant side are accounted for by the product side, you must balance the chemical
equation. The chemical equation gives information regarding the nature of the reactants and products and
the relative numbers of each. By specifying the compounds in the equation, we know the physical states we
are dealing with. The relative numbers of reactants and products are indicated by coefficients.
3.7: Balancing Chemical Equations
The same number of each type of atom must be found among the reactants and products. The identity of the
reactants and products of a reaction are determined by experimental observation. The formulas of the
compounds must never be changed in balancing a chemical equation.
3.8: Stoichiometric Calculations
The first step is to balance the equation. Next, convert the known mass of the reactant or product to moles
of that substance. Then use the balanced equation to set up appropriate mole ratios. After, use the
appropriate mole ratios to calculate the number of moles of the desired reactant or product. Last, convert
from moles to grams.
3.9: Calculations Involving a Limiting Reactant
In some reactions, one compound gets consumed faster than others. If there are excess and limits to the
amount of compound, you must find which compound will be used up first. Use stoichiometry to find
moles of each compound. Compare with that of the actual mole ratio. To find percent yield, divide actual
yield by theoretical yield and multiply by 100%.
II.
1. When a sample of natural copper is vaporized and injected into a mass spectrometer, the mass values for
Copper 63 and Copper 65 are 62.93 amu and 64.93 amu, respectively. Use these data to compute the
average mass of natural copper.
a)
b)
c)
d)
64.72 amu
63.55 amu
69.24 amu
74.31 amu
2. Calculate the number of moles in a sample of cobalt containing 5.00 x 10^20 atoms.
a)
b)
c)
d)
8.20 x 10^ -4
8.30 x 10^ -4
8.73 x 10^ -5
9.42 x 10^ -8
3. Calculate the mass of the given amount of cobalt from the previous question.
a)
b)
c)
d)
5.24 x 10^ -3
5.83 x 10^ 2
7.59 x 10^ -4
4.89 x 10^ -2
4. A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass. The
compound has a molar mass of 283.88 g/mol. What is the compound’s empirical formula?
a)
b)
c)
d)
P2O5
P3O4
PO3
P2O6
5. Find the molecular formula from the question above.
a)
b)
c)
d)
P6O10
P2O5
P8O12
P4O10
III.
Given the compound C10H6O3
a)
Calculate the molar mass.
b) A sample of 1.56 x 10-2 g of the compound was extracted. How many moles does it represent?
Work
II
1. (69.09 atoms)(62.93 amu/atom) + (30.91 atoms)(64.93 amu/atom) = 6355 amu
Need to find average mass: (6355 amu/100 atoms) = 63.55 amu/atom
2. 5.00 x 10^ 20 atoms Co x (1 mol CO/ 6.022 x 10^23 atoms C) = 8.30 x 10^-4 mol Co
3. 8.30 x 10^ -4 mol Co x (58.93 g Co/ 1 mol Co) = 4.89 x 10^-2 g Co
4. Find the Empirical Formula
43.64 g P x (1 mol P/ 30.97 g P) = 1.409 mol P
56.36 g O x (1 mol O/ 16.00 g O) = 3.523 mol O
1.409 mol / 1.409 mol = 1
3.523 mol / 1.409 mol = 2.5
= P2O5
5. Find the Molecular Formula
Need (Molar Mass / Empirical formula mass)
283.88 g / 141.94 g = 2
2(P2O5)
= P4O10
III
a. 10 C: 10 x 12.01 g = 120.1 g
6 H: 6 x 1.008 g = 6.048 g
3 O: 3 x 16.00 g = 48.00 g
Mass of C10H6O3 = 174.1 g
b. 1.56 x 10^ -2 g C10H6O3 x ( 1 mol C10H6O3 / 174.1 C10H6O3) = 8.96 x 10^-5 mol C10H6O3