Chemistry - Chemical Calculations

[Concept] What is formula mass?

The sum of the atomic masses of every atom shown in a chemical formula.

[Rule] How do you calculate formula mass?

Multiply each element's atomic mass by its subscript, then add all contributions.

[Concept] Formula mass versus molecular mass

Formula mass is commonly used for ionic compounds; molecular mass is commonly used for molecular compounds. The calculation is the same.

[Practice] Calculate the formula mass of MnCO3 using Mn 54.936, C 12.011, and O 15.9994.

54.936 + 12.011 + 3(15.9994) = 114.945 amu.

[Concept] What is a mole?

A counting unit equal to 6.02 x 10^23 specified particles.

[Concept] What is Avogadro's number?

6.02 x 10^23 particles per mole.

[Rule] What may be counted with moles?

Atoms, molecules, formula units, ions, or any other specified chemical entities.

[Rule] Particles to moles

Divide by 6.02 x 10^23 particles/mol.

[Rule] Moles to particles

Multiply by 6.02 x 10^23 particles/mol.

[Practice] How many moles are in 9.16 x 10^25 atoms?

9.16 x 10^25 / 6.02 x 10^23 = 1.52 x 10^2 mol.

[Practice] How many molecules are in 0.39 mol?

0.39 x 6.02 x 10^23 = 2.3 x 10^23 molecules.

[Concept] What is molar mass?

The mass in grams of one mole; it is numerically equal to formula mass in amu.

[Rule] Grams to moles

Divide mass by molar mass.

[Rule] Moles to grams

Multiply moles by molar mass.

[Rule] Central conversion map

Particles

[Practice] How many moles are in 36.0 g H2O?

36.0 g / 18.02 g/mol = 2.00 mol H2O.

[Practice] What mass is 0.250 mol CO2?

0.250 mol x 44.01 g/mol = 11.0 g CO2.

[Practice] How many molecules are in 2.00 mol H2O?

2.00 x 6.02 x 10^23 = 1.20 x 10^24 molecules.

[Practice] How many atoms are in 425 g Ag if Ag is 107.868 g/mol?

425/107.868 x 6.02 x 10^23 = 2.37 x 10^24 atoms.

[Rule] How do formula subscripts become mole relationships?

One mole of a compound contains the subscript number of moles of each element.

[Practice] What amounts of atoms are contained in 1 mol C7H8N4O2?

7 mol C, 8 mol H, 4 mol N, and 2 mol O atoms.

[Practice] How many grams of carbon are in 0.300 g C7H8N4O2?

0.300 g x (7 x 12.011 g C / 180.17 g compound) = 0.140 g C.

[Concept] What is a chemical equation?

A symbolic statement showing the reactants, products, and changes in a chemical reaction.

[Rule] Where are reactants and products written?

Reactants are on the left; products are on the right; the arrow means yields.

[Memory] Physical-state symbols

s = solid; l = liquid; g = gas; aq = dissolved in water.

[Rule] What two requirements must a valid equation meet?

It must represent a reaction that can occur and it must conserve every type of atom.

[Concept] Law of conservation of mass

Atoms are not created or destroyed in a chemical reaction, so total mass remains constant.

[Concept] What is a balanced equation?

An equation with the same number of each kind of atom on both sides.

[Rule] What may be changed when balancing?

Only coefficients placed before formulas.

[Trap] Why must subscripts never be changed to balance an equation?

Changing a subscript changes the identity of the substance.

[Rule] Balancing workflow

Count atoms, choose a useful element to balance first, adjust coefficients one element at a time, recheck all atoms, and reduce to the smallest whole numbers.

[Practice] Balance Fe + O2 -> Fe2O3.

4 Fe + 3 O2 -> 2 Fe2O3.

[Practice] Balance NaN3 -> Na + N2.

2 NaN3 -> 2 Na + 3 N2.

[Practice] Balance H2 + O2 -> H2O.

2 H2 + O2 -> 2 H2O.

[Practice] Balance UO2 + HF -> UF4 + H2O.

UO2 + 4 HF -> UF4 + 2 H2O.

[Practice] Balance C2H8N2 + N2O4 -> N2 + CO2 + H2O.

C2H8N2 + 2 N2O4 -> 3 N2 + 2 CO2 + 4 H2O.

[Concept] What quantitative information do balanced coefficients provide?

Relative numbers of particles and mole ratios among every reactant and product.

[Concept] What is a mole ratio?

A conversion factor made from coefficients in a balanced equation.

[Trap] Why must the equation be balanced before stoichiometry?

Only balanced coefficients give valid conserved mole relationships.

[Rule] Standard stoichiometry path

Given unit -> moles of given -> moles of wanted using a coefficient ratio -> wanted unit.

[Rule] What converts between different substances?

A mole ratio from the balanced equation.

[Rule] What converts between grams and moles of one substance?

That substance's molar mass.

[Practice] For CH4 + 2 O2 -> CO2 + 2 H2O, give the CH4:H2O ratio.

1 mol CH4 : 2 mol H2O.

[Practice] For the same equation, how many moles H2O form from 3.00 mol CH4?

3.00 x 2/1 = 6.00 mol H2O.

[Practice] How many grams H2O form from 15.0 g CH4 with excess O2?

15.0/16.04 x 2 x 18.015 = 33.7 g H2O.

[Practice] For N2 + 3 H2 -> 2 NH3, how many moles NH3 form from 4.50 mol H2?

4.50 x 2/3 = 3.00 mol NH3.

[Practice] For N2 + 3 H2 -> 2 NH3, how many grams NH3 form from 1.00 mol N2?

1.00 x 2 x 17.03 = 34.1 g NH3.

[Practice] For N2H4 + 2 H2O2 -> N2 + 4 H2O, what is the H2O2:N2H4 ratio?

2 mol H2O2 : 1 mol N2H4.

[Practice] What mass N2H4 reacts with 1.25 kg H2O2?

1250/34.014 x 1/2 x 32.046 = 589 g N2H4.

[Concept] What is theoretical yield?

The maximum product amount predicted by stoichiometry if no losses or inefficiencies occur.

[Concept] What is actual yield?

The amount of product actually collected in an experiment.

[Rule] Percent-yield equation

Percent yield = actual yield / theoretical yield x 100%.

[Practice] Theoretical yield is 25.0 g and actual yield is 20.0 g. Find percent yield.

20.0/25.0 x 100% = 80.0%.

[Practice] A reaction has 85.0% yield and theoretical yield 40.0 g. Find actual yield.

0.850 x 40.0 = 34.0 g.

[Practice] Actual yield is 12.5 g and percent yield is 62.5%. Find theoretical yield.

12.5/0.625 = 20.0 g.

[Concept] Why is actual yield often below theoretical yield?

Incomplete reaction, side reactions, product loss, or measurement error.

[Practice] For 3 Ca + N2 -> Ca3N2, 29.0 g N2 gives a theoretical yield of about what mass Ca3N2?

29.0/28.014 x 148.248 = about 153 g Ca3N2.

[Practice] If 120.3 g Ca3N2 is obtained from that reaction, find percent yield.

120.3/153.4 x 100% = 78.4%.

[Concept] What is the limiting reactant?

The reactant used up first; it determines the maximum product amount.

[Concept] What is an excess reactant?

A reactant present in more than the required amount, so some remains after the limiting reactant is gone.

[Rule] Product method for finding the limiting reactant

Calculate how much of the same product each reactant could make; the reactant producing less product is limiting.

[Rule] Needed-amount method for finding the limiting reactant

Use the balanced ratio to calculate how much of one reactant is required for the other and compare required with available.

[Trap] Can you compare reactant masses directly to find the limiting reactant?

No. Convert to moles and use the balanced mole ratio.

[Practice] For 3 H2 + N2 -> 2 NH3, 5.0 mol H2 reacts with 1.5 mol N2. Which is limiting?

N2 is limiting; it needs 4.5 mol H2, and 5.0 mol is available.

[Practice] In that Haber-process mixture, how many moles NH3 form?

1.5 mol N2 x 2 = 3.0 mol NH3.

[Practice] In that mixture, how many moles H2 remain?

5.0 - 4.5 = 0.5 mol H2.

[Practice] For 2 H2 + O2 -> 2 H2O, 4.00 mol H2 reacts with 1.50 mol O2. Which is limiting?

O2 is limiting; it consumes 3.00 mol H2.

[Practice] How much H2O forms in that mixture?

1.50 mol O2 x 2 = 3.00 mol H2O.

[Practice] For C2H4 + HCl -> C2H5Cl, convert 8.00 g C2H4 and 12.0 g HCl to moles.

C2H4: 0.285 mol; HCl: 0.329 mol.

[Practice] Which reactant limits the ethyl-chloride reaction?

C2H4 is limiting because the reaction is 1:1 and it has fewer moles.

[Practice] What theoretical mass C2H5Cl forms from 8.00 g C2H4?

0.285 mol x 64.512 g/mol = 18.4 g.

[Practice] If 10.6 g C2H5Cl forms, what is percent yield?

10.6/18.4 x 100% = 57.6%.

[Rule] Which reactant must be used to calculate theoretical yield?

Always the limiting reactant.

[Rule] Full limiting-reactant/yield workflow

Balance; convert each reactant to moles; identify the limiting reactant; calculate theoretical product; compare actual and theoretical yields.

[Trap] What are three common stoichiometry mistakes?

Using an unbalanced equation, skipping the mole ratio, and using the excess reactant to calculate theoretical yield.