MASS RELATIONSHIPS IN CHEMICAL REACTIONS
This chapter reviews the mole concept, balancing chemical equations, and stoichiometry. The topics covered in this chapter are:
• Atomic mass and average atomic mass
• A vogadro’ s number, mole, and molar mass
• Percent composition calculations
• Empirical and molecular formula determinations
• Chemical equations, amount of reactant and product calculations
• Limiting reagents and reaction yield calculations
Take Note: It is absolutely essential that you master the mole concept to do well on the quantitative aspects of AP Chemistry!!
When solving quantitative problems on the Free Response section of the AP exam, supporting work ...view middle of the document...
00000 amu and 13.00335 amu, respectively.)
- 24 -
Solution to Example 1.
(12.00 amu × 0.9890) + (13.00335 amu × 0.0110) = 12.01 = average atomic mass of carbon
Molecular mass refers to the mass of a molecule expressed in atomic mass units. The molecular mass is the sum of the atomic masses of the atoms in the molecular formula.
Example 2. Determining molecular mass.
Determine the molecular mass of barium hydroxide, Ba(OH)2 .
Solution to Example 2.
1 Ba atom + 2 O atoms + 2 H atoms
137.33 amu + 2(16.00 amu) + 2(1.0079 amu) = 171.34 amu
Avogadro’s number, mole, and molar mass
Avogadro’s number refers to 6.022 × 1023 particles. The quantity of an element that contains Avogadro’s number of particles is called a mole. The molar mass of an element or compound is the mass of one mole of its atoms or molecules expressed in grams.
Some examples of mole, atom, ion, molecule, and mass equivalencies are: • 1moleofCatoms= 6.022×1023 atomsofC= 12.01gofC
• 1 mole of O2 molecules = 6.022 × 1023 molecules of O2 = 32.00 g O2
• 1 mole of of O2 molecules = 2 moles of oxygen atoms = 12.044 × 1023 atoms of O
• 1 mole O2– ions = 6.022 × 1023 O2– ions = 16.00 g O2– ions
Percent composition calculations
The percent composition of an element in a compound =
n × molar mass of element × 100% where molar mass of compound
n is the number of moles of the element in the compound
￼- 25 -
Example 3. Determining the percent composition.
Determine the percent composition of chloride in barium chloride, BaCl2.
Mass of 2 Cl ̄ × 100 = 2(35.45 g Cl ̄)_____
Mass of BaCl2 2(35.45 g Cl ̄) + 137.33 g Ba × 100% = 34.0 % Cl ̄
Empirical and molecular formula determinations
The empirical formula is the most reduced form of the molecular formula. For instance, sugar, which has the molecular formula of C6H12O6, has an empirical formula of CH2O. Empirical formulas can be determined from percent composition data. If the molar mass is also known, the molecular formula can be determined as well (see Example 4).
Example 4. Determining empirical and molecular formulas.
An unknown compound is known to contain 30.43% N and 60.56% O and has a molar mass of 92.00 g. What are the empirical formula and the molecular formula for the compound?
￼￼￼￼Solution to Example 4
￼￼￼Assume a 100 g sample.
￼￼￼30.43 g of N 60.56 g O
￼￼￼Convert gram quantities to moles by dividing by the molar mass of the atom.
￼￼￼30.43gN _ 14.01 g N/mol N
60.56 g O 16.00 g O/ mol O
= 2.17 mol N = 4.35 mol
￼￼￼￼￼Divide by the smallest number of moles to obtain whole number ratios. The whole number ratios are the subscripts in the empirical formula.
￼￼￼2.17 mol N 2.17 mol N
4.35 mol O 2.17 mol N
= 1molN = 2mol
￼￼￼￼￼To determine the molecular formula divide the molar mass by the empirical mass. Multiply the subscripts of the empirical formula by this factor.
￼￼Empirical mass of NO2 = 14.01 g N + 2(16.00) g O = 46.01 g NO2
Molar mass = 92.00 g = 2...