Chapter 3 Review: Mass Relationships in Chemical Reactions
Will
1. 1g=1AMU
2. Calculate the average atomic mass given the mass and natural abundance of each
isotope:
Element: Copper
Isotopes and % Abundance: Cu-63 (69.17%), Cu-65 (30.08%)
Calculations: 63(.6917)+65(.3008)= average atomic mass
(43.58)+(20.04)= 63.62 g
3. Avogadro’s Number: 6.022e23 atoms per mol
4. Determine the number of atoms present given the number of moles:
For example, if you had .647 mol Carbon
(.647)(6.022e23)= 3.90e23 atoms
5. Write the conversion of mass, number of moles, and number of atoms for an
element/compound.
MassMole:
(grams given)(1 mol of element/compound)/(molar mass of element/compound)
Moles Atoms:
(given # of moles)(6.022e23)
MassAtoms:
(grams given)(1 mol)(6.022e23)/(molar mass)(1 mol)
6. Calculate the molecular mass and molar mass given the molecular formula
Compound: Glucose
Molecular Formula: C6H12O6
Molecular Mass: 6(12.01)+12(1.008)+6(16.00)= 180.156g
Empirical Formula: CH2O
Molar Mass: (12.01)+2(1.008)+(16.00)= 30.026g
7.
The invention of the mass spectrometer led to the discovery of isotopes. There are 5
steps to the mass spectrometer. Step 1: Inject and Vaporize- Atoms separate from each
other. Step 2: Ionize- electron gun bombards the atoms with electrons. Step 3:
Acceleration- an electric field shoots the atoms through the spectrometer. Step 4:
Deflection: in a magnetic field, heavier isotopes get deflected less than lighter isotopes.
Step 5: Detection: counts how many of each mass come through by using electric of
photographic methods.
This is a mass spectrograph. It shows the percent abundance of each isotope. As you can see,
the 35g isotope is much more abundant than the 37g isotope. To determine the percentage of
each isotope from a graph, you must measure each of the lines in centimeters. Then, for each
line you put the length of that line divided by the total then multiplied by 100 in order to put it
into percent form.
8. Percent composition of elements in a compound:
(Molar mass of element)/(total mass of compound) x100= % Composition
Compound AgNO3
Molar Mass: 169.87
Elements: Ag- 107.87/169.87= .635 x100= 63.5%
N: 14.00/169.87= .0824 x100= 8.24%
O: 3(16.00)/169.87= .283 x100= 28.3%
9. Determining Empirical Formula:
The best way to remember how to do empirical formula is the helpful rhyme: “Mass to
mole, then divide by small, multiply ‘til whole”.
For Example:
What is the empirical formula for a brown gas known to contain 2.34 g N and 5.34 g O?
N= 2.34g/14.01g=.167/.167= 1
O= 5.34g/16.00g=.334/.167=2
Therefore the empirical formula would be NO2
Peter
10. Calculate the molecular formula given the mass, or mass percent, of each element present
 Numbers will not be whole because they are based on average mass of the common isotopes
 Empirical means the lowest ratio, molecular means the actual ratio that exists.

To calculate: Divide the Molecular formula mass by the empirical formula mass to get a whole number.
Then, multiply the whole number to the subscripts of the empirical formula
11. Balance Chemical Equations
 Chemical Equation: Sentence that describes what happens in a chemical reaction.
 Law of conservation of mass: Matter can neither be created nor destroyed, just changed in form.
 Equations need to be balanced in order to satisfy this law.
 Balancing tips: Never start with (H) or (O). Treat polyatomic ions as chunks not separate atoms.
12. Interpret the meaning of chemical equations in terms of molecules, moles, and masses
 The coefficients of a chemical equation can represent the relationship of molecules or moles.
 These coefficients DO NOT in any way show mass relationships.
13. Distinguish between products and reactants in a chemical equation
 The products are on the right side of the chemical equation. These are what the reactants react to form.
 Reactants are on the left side. These substances are what you start with.
14. Predict the products formed by combustion reactions
 In combustion reactions the products formed are always H2O and CO2.
 After writing these two products it is important that the equation is then balanced.
15. Use Stoichiometric methods to predict the mass of products formed given the mass of each reactant
 Steps: Mole –mole: Use the mole ratio to convert from moles of substance A to moles of substance B
Mole –mass: Use the moles of substance B obtained from the previous step and then multiply it by
the molar mass of the substance in order to get grams of B.
Mass –mole: Using the answer in the previous step, set up a conversion factor with the mole ratio in
order to get moles of substance B.
Mass –mass: Convert Mass of A to moles of A then using the mole ration convert moles of A to
Moles of B. Then using a conversion factor, convert moles of B to mass of B.
16. Determine limiting and excess reagents. Amount of reagents left over
 Determine LR steps: Take reactants and convert to moles. Divide each mole amount by its coefficient. The
smaller number is the LR. Perform stoich. calc. w/ LR to find theoretical yield.
 In order to determine the amount of excess reagent use the original amount of reactant to perform a stoich
calculation. Then, perform the same calculation w/ the LR. Finally, Subtract the 2 number.
17. Predict the Theoretical and actual yield of a reaction using stoichiometric calculations
 The theoretical yield is the numerical value obtained by performing a stoichiometric calculation with the
LR, if there is one.
 The actual yield is the amount of product physically produced when you perform the experiment or allow
the reaction to go to completion.
18. Calculate the mass of each reactant given the percent yield and the mass of products required
 Percent yield: Ratio in (%) of amount of product produced in lab compared to the amount of product
theoretically produced determined through a stoichiometric calculation.
 Theoretical yield: Amount you would produce in lab if everything went perfect in lab and there were no
errors.
 Actual yield: What you make in lab with imperfect conditions.
 (%) yield = (actual)/(theoretical) x 100
Multiple choice questions
1. What is the mass percent of (N) in a 28g sample of NaNO 3?
a) 974.7654321%
c) 7.61%
b) 5.43%
d) 4.35%
N- 14.01/85 = 16.5% N per mole of NaNO3
28g/85g = .329 mol NaNO3
NaNO3
.165 x .329 = .0543(100) 5.43% N in 28g of NaNO3
To find mass percent, you divide the molar mass of the element by the total mass of the compound. Then, multiply
that number by the number of moles. Multiply that number by 100 to get it into a percent.
2. Which of the following statements are true?
I. Coefficients in a balanced chemical equations can be used to show mass relationships.
II. Coefficients in a balanced chemical equations can be used to show mole relationships.
III. Coefficients in a balanced chemical equations can be used to show molecule relationships.
IV. 6.022x1023 is equal to Avogadro’s number.
a) All are true.
c) I, II, II only.
b) None are true.
d) II, III, IV only.
Number IV is a known fact. I is incorrect because all element have different masses therefore, the coefficient cannot
relate these numbers. Moles and molecules are related through Avogadro’s number therefore moles and molecules
can be related though the coefficients in a balanced chemical equation.