Post on 29-Dec-2015
transcript
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Chapter 3
Stoichiometry: Calculations with Chemical Formulas and Equations
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Chemical Equations
2 H2 + O2 → 2 H2O
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Chemical Equations
CH4 + O2 → CO2 + H2O
CH4 + 2 O2 → CO2 + 2 H2O
Before:
After:
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Formula Weights
H2SO4 = 2(1.01) + 32.06 + 4(16.00)
H2SO4 = 98.1 amu
% element = (number of atoms)(atomic mass) x 100%
formula weight of compound
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Avogadro's Number and the Mole
The number of particles in a mole is called Avogadro’s number, which is 6.02 x 1023.
1 mol carbon atoms = 6.02 x 1023 carbon atoms1 mol H2O molec. = 6.02 x 1023 H2O molec.
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Avogadro’s Number and the Mole
formula weight of H2SO4 = 98.1 amu
molar mass of H2SO4 = 98.1 g/mol
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Avogadro’s Number and the Mole
grams ↔ moles: use molar mass moles ↔ particles: use Avogadro’s number
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Empirical Formulas from Analyses
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Empirical Formula from Analyses
Ascorbic acid (vitamin C) contains 40.92% C, 4.59% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?
1) Assume a 100 g sample and change % to g:40.92% C = 40.92 g C4.59% H = 4.59 g H54.50% O = 54.50 g O
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Empirical Formula from Analyses
2) Convert grams of each element into moles:
40.92 g C x 1 mol C = 3.407 mol C
12.01 g C
4.59 g H x 1 mol H = 4.54 mol H
1.008 g H
54.50 g O x 1 mol O = 3.406 mol O
16.00 g O
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Empirical Formula from Analyses
3) Calculate the mole ratio for each element by dividing each mol value by the smallest mol value:
C : 3.407 = 1.000 3.406
H : 4.54 = 1.33 3.406
O: 3.406 = 1.000 3.406
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Empirical Formula from Analyses
The ratio for H is too far from 1 for us to round, so we need to multiply each number by 3:
C : H : O = 1 : 1.33 : 1
C : H : O = 3(1 : 1.33 : 1) = 3 : 4 : 3
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Empirical Formula from Analyses
The whole number ratio gives us the subscript values for the empirical formula:
C3H4O3
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Empirical Formula from Analyses
On Your Own:
A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, contains 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?
C4H4O
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Empirical Formula from Analyses
Combustion analysis is used to calculate empirical formulas for compounds containing carbon and hydrogen.
The amounts of CO2 and H2O produced give the moles of H and C in the original compound.
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Molecular Formula from Empirical Formula
The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula
Whole-number multiple = molecular weight , empirical formula weight
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Quantitative Information from Balanced Equations
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Stoichiometric Calculations
How many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6?
C6H12O6(s)+ 6 O2(g)→ 6 CO2(g)+ 6 H2O(l)
1) Convert grams of glucose into moles of glucose:
1.00 g C6H12O6 x 1 mol = 5.56 x 10-3 mol C6H12O6
180.0 g
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Stoichiometric Calculations
2) Convert moles of glucose into moles of water using the ratio from the chemical equation:
5.56 x 10-3 mol C6H12O6 x 6 mol H2O = 3.33 x 10-2 mol H2O
1 mol C6H12O6
3) Convert moles of water into grams of water:
3.33 x 10-2 mol H2O x 18.0 g H2O = 0.600 g H2O
1 mol H2O
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Limiting Reactants
N2 (g) + 3 H2 (g) → 2 NH3 (g)
How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?
3.0 mol N2 x (3 mol H2) = 9.0 mol H2
(1 mol N2)
6.0 mol H2 x (2 mol NH3) = 4.0 mol NH3
(3 mol H2)
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Theoretical Yield
The quantity of product that is calculated to form when all of the limiting reactant reacts is called the theoretical yield.
The quantity of product that is actually obtained in a reaction is called the actual yield.
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Theoretical Yield
Percent yield = actual yield x 100% theoretical yield