Chapter 3 - Stoichiometry Flashcards
-Balance chemical equations -Calculate molecular weights -Convert grams --> moles & moles --> grams using molar mass -Convert # of molecules --> moles & vice versa using Avogadro's # -Calculate the empirical & molecular formula of a compound from % composition & molecular weight -Calculate amounts, in grams or moles, of reactants & products for a reaction -Calculate the % yield of a reaction
3.2
Write the equation for a Combination reaction.
A + B —-> C
3.2
Write the equation for a Decomposition reaction.
C —–> A + B
3.2
Write the equation for a Combustion reaction.
Fuel + O2(g) —> CO2(g) + H2O(l)
3.3
Define “formula weight”.
The sum of the atomic weights of each atom in its chemical formula.
Ex. H2SO4 = 2(1.0 amu) + 32.1 amu + 4(16.0 amu) = 98.1 amu
3.3
Define “molecular weight”.
The mass of the collection of atoms represented by the chemical formula for a molecule.
Ex. C6H12O6 = 6(12.0 amu) + 12(1.0 amu) + 6(16.0 amu) = 180.0 amu
3.3
Write the equation for % element in a substance.
% element= ((# atoms of said element * atomic weight of element) / formula weight of compound) x 100
3.3 Samp. Ex
Calculate the % of C, H, & O (by mass) in C12H22O11.
% C = ((12 * 12.01 amu)/342.22 amu) x 100 = 42.1%
% H = ((22 * 1.01 amu)/342.22 amu) x 100 = 6.4%
% O = ((11 * 16 amu)/342.22 amu) x 100 = 51.5%
3.4
1 mol NO3- ions = x ions
6.022 x 10^23
3.4 Samp. Ex.
Without using a calculator, arrange the following samples in order of increasing numbers of C atoms: 12g C-12, 1 mol C2H2, 9 x 10^23 molecules of CO2.
12g C-12 (6 x 10^23 atoms) < 9 x 10^23 molecules of CO2 (9 x 10^23 atoms because each molecule has 1 C atom) < 1 mol C2H2 (~12 x 10^23 atoms)
3.4 Practice Ex.
Arrange the following samples in order of increasing # of O atoms: 1 mol H2O, 1 mol CO2, 3 x 10^23 molecules O3
1 mol H2O (6 x 10^23 atoms) < 3 x 10^23 molecules O3 (9 x 10^23 atoms) < 1 mol CO2 (12 x 10^23 atoms)
3.4
Define “molar mass”.
The mass in grams of 1 mol of a substance.
The molar mass (in g/mol) of any substance is always numerically equal to its formula weight (in amu).
3.5 Samp. Ex.
Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is its empirical formula?
- 92g (1 mol/12.01g) = 3.407 mol C
- 58g (1 mol/1.01g) = 4.54 mol H
- 50g (1 mol/16g) = 3.406 mol O
Divide them all by 3.407, the smallest number.
C = 1, H = 1.33, O = 1
Multiply by 3 to get whole numbers.
C= 3, H = 4, O = 3
C3H4O3
3.5 Prac. Ex.
A 5.325g sample of methyl benzoate contains 3.758g C, 0.316 g H, and 1.251g O. What is its empirical formula?
- 758g C (1 mol / 12.01g) = .3129 mol C
- 316g H (1 mol / 1.01g) = .313 mol H
- 251g O (1 mol / 16g) = .0782 mol O
Divide by .0782.
C: 4, H: 4, O: 1
C4H4O
3.5
Whole-number multiple = ? / ?
Molecular weight/empirical formula weight
3.5 Samp. Ex
Mesitylene has an empirical formula of C3H4. The molecular weight is 121 amu. What’s the molecular formula?
121/40.07 = 3
Multiply by 3.
C9H12
3.5 Samp. Ex.
Isopropyl alcohol is composed of C, H, and O. Combustion of 0.255g of isopropyl alcohol produces .561g of CO2 & .306g of H2O. Determine isopropyl alcohol’s empirical formula.
.561g CO2 (1 mol / 44.01g)(1 mol C/ 1 mol CO2)(12.01g C / 1 mol C) = 0.153g C
.306g H2O (1 mol / 18.02g)(2 mol H / 1 mol H2O)(1.01g H / 1 mol H) = 0.0343g H
Mass of O = .255g isopropyl alcohol - 0.153g C - 0.0343g H = .0677g O
Gram to mol conversions.
0.0128 mol C, 0.0340 mol H, 0.0043 mol O.
Divide by .0043 to produce 3:8:1 ratio.
C3H8O
3.6 Give It Some Thought (p. 99)
When 1.57 mol O2 reacts with H2 to form H2O, how many moles of H2 are consumed in the process?
2 H2 + O2 –> 2 H2O
1.57 mol O2 (2 mol H2 / 1 mol O2) = 3.14 mol H2
3.7
How do you find the limiting reactant?
Figure out which reactant produces the smallest amount of product. That’s the limiter.
ex. 150g H2 (1 mol / 2.02g)(2 mol H2O / 1 mol H2)(18.02g H2O / 1 mol H2) = 74g H2O
1500g O2 (1 mol / 32g)(2 mol H2O / 1 mol O2)(18.02g H2O / 1 mol) = 1300g H2O
Hydrogen limits.
3.7
Define “theoretical yield”.
The quantity of product that’s calculated to form when all of the limiting reactant reacts.
3.7
Define “actual yield”.
The amount of product actually obtained in a eraction.
3.7
Percent yield = (? / ?) x 100
Find the percent yield of adipic acid. The theoretical yield is 43.5g. The actual yield is 33.5g.
(Actual / theoretical) x 100
(33.5g / 43.5g) x 100 = 77.0%
3.1 - Give It Some Thought (p. 80)
How many atoms of Mg, O, and H are represented by 3 Mg(OH)2?
3 atoms Mg, 6 atoms O, 6 atoms H
