 # 1 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations.

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1 Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

2 Chemical Equations 2 H 2 + O 2 → 2 H 2 O

3 Chemical Equations CH 4 + O 2 → CO 2 + H 2 O CH 4 + 2 O 2 → CO 2 + 2 H 2 O Before: After:

4 Formula Weights H 2 SO 4 = 2(1.01) + 32.06 + 4(16.00) H 2 SO 4 = 98.1 amu % element = (number of atoms)(atomic mass) x 100% formula weight of compound

5 Avogadro's Number and the Mole The number of particles in a mole is called Avogadro’s number, which is 6.02 x 10 23. 1 mol carbon atoms = 6.02 x 10 23 carbon atoms 1 mol H 2 O molec. = 6.02 x 10 23 H 2 O molec.

6 Avogadro’s Number and the Mole formula weight of H 2 SO 4 = 98.1 amu molar mass of H 2 SO 4 = 98.1 g/mol

7 Avogadro’s Number and the Mole grams ↔ moles: use molar mass moles ↔ particles: use Avogadro’s number

8 Empirical Formulas from Analyses

9 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 C 4.59% H = 4.59 g H 54.50% O = 54.50 g O

10 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

11 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

12 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

13 Empirical Formula from Analyses The whole number ratio gives us the subscript values for the empirical formula: C3H4O3C3H4O3

14 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? C4H4OC4H4O

15 Empirical Formula from Analyses Combustion analysis is used to calculate empirical formulas for compounds containing carbon and hydrogen. The amounts of CO 2 and H 2 O produced give the moles of H and C in the original compound.

16 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

17 Quantitative Information from Balanced Equations

18 Stoichiometric Calculations How many grams of water are produced in the oxidation of 1.00 g of glucose, C 6 H 12 O 6 ? C 6 H 12 O 6 (s)+ 6 O 2 (g)→ 6 CO 2 (g)+ 6 H 2 O(l) 1) Convert grams of glucose into moles of glucose: 1.00 g C 6 H 12 O 6 x 1 mol = 5.56 x 10 -3 mol C 6 H 12 O 6 180.0 g

19 Stoichiometric Calculations 2) Convert moles of glucose into moles of water using the ratio from the chemical equation: 5.56 x 10 -3 mol C 6 H 12 O 6 x 6 mol H 2 O = 3.33 x 10 -2 mol H 2 O 1 mol C 6 H 12 O 6 3) Convert moles of water into grams of water: 3.33 x 10 -2 mol H 2 O x 18.0 g H 2 O = 0.600 g H 2 O 1 mol H 2 O

20 Limiting Reactants N 2 (g) + 3 H 2 (g) → 2 NH 3 (g) How many moles of NH 3 can be formed from 3.0 mol of N 2 and 6.0 mol of H 2 ? 3.0 mol N 2 x (3 mol H 2 ) = 9.0 mol H 2 (1 mol N 2 ) 6.0 mol H 2 x (2 mol NH 3 ) = 4.0 mol NH 3 (3 mol H 2 )

21 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.

22 Theoretical Yield Percent yield = actual yield x 100% theoretical yield

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