1. Stoichiometry Chapter 3 E-mail: benzene4president@gmail.com
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2. Stoichiometry - Ch. 3
1. For a new element, 67.16% is an isotope with mass amu, 2.76% is an isotope with mass amu and 30.08% is an isotope with mass amu. Calculate the average atomic mass for this new element?

3. Stoichiometry - Ch. 3 Average Atomic Mass =
(fraction of isotope A)(mass of isotope A)
+ (fraction of isotope B)(mass of isotope B) + etc.

4. Stoichiometry - Ch. 3
2. For which of the following compounds does 1.00 g represent 3.32 × 10-2 mol?
a. NO2
b. H2O
c. C2H6
d. NH3
e. CO

5. Stoichiometry - Ch. 3 1 mole of PCl5 208.239 g of PCl5
5(35.453) g of Cl g
of P
1 mole of
PCl5
5 moles of Cl P
6.022x1023
molecules
6(6.022x1023) total atoms
P atoms
5(6.022x1023)Cl atoms

6. Stoichiometry - Ch. 3
3. If a sample of diatomic element weighs g and contains 4.162x1024 atoms. Identify the element.

7. Stoichiometry - Ch. 3 4. If you have 0.63 mg of H2SO4
a. How many H2SO4 molecules are in your sample?
b. How many oxygen atoms?

8. Stoichiometry - Ch. 3
5. An alkali metal oxide contains 83.01% metal by mass. Determine the identity of the metal oxide. How many grams of oxygen are in a 25.0 g sample of the metal oxide?

9. Percent by mass = (mass of 1 part)x100 (Total mass)
Stoichiometry - Ch. 3
Percent by mass = (mass of 1 part)x100 (Total mass)

10. Stoichiometry - Ch. 3
6. Compound X2Y is 60% X by mass. Calculate the percent Y by mass of the compound XY3?

11. Stoichiometry - Ch. 3
7. Tryptophan is 64.7% carbon, 5.9% hydrogen, 13.7% nitrogen and 15.7% oxygen. What is the empirical formula for tryptophan?

12. Stoichiometry - Ch. 3
Empirical Formula ⇒ The lowest whole number molar ratio of the elements in a compound
1. Convert given values into moles
2. Divide all moles by the smallest mole value
3. If you have all whole numbers you have the EF – if not try multiplying them all by 2 or 3 etc.

13. Stoichiometry - Ch. 3
8. The empirical formula for xylene is C4H5 and xylene has a molar mass of g/mol. Determine the molecular formula for xylene.

14. Stoichiometry - Ch. 3
Molecular Formula ⇒ The actual molar ratio of the elements in a compound – it is some multiple of the empirical formula (x1, x2 etc)
1. Derive empirical formula
2. Determine the empirical mass
3. (Molar mass)/(empirical mass) = multiple
4. Multiply the empirical formula by the multiple

15. Stoichiometry - Ch. 3 9. Consider the following unbalanced reaction:
NH3 + O2  NO2 + H2O
a. How many moles of oxygen gas are required to make moles of nitrogen dioxide?
b. How many grams of water can be produced from 9.64 g of ammonia?
c. Identify the limiting reagent if 3 moles of ammonia is combined with 5 moles of oxygen
d. Identify the limiting reagent if 10 g of ammonia is combined with 28 g of oxygen

16. Methodology for Reaction Stoichiometry Problems
Stoichiometry - Ch. 3
Methodology for Reaction Stoichiometry Problems
1. Write a balanced chemical reaction
2. Convert given value(s) into moles (you may have to ID the limiting reagent – next slide)
3. Use reaction coefficients as a molar ratio
4. Convert moles of your unknown into the desired units

17. Stoichiometry - Ch. 3 Limiting Reagent ⇒ Limits the amount of
product that is produced due to running out 1st
The limiting reagent is used to determine the
maximum yield of product/s aka the theoretical
yield and the maximum consumption of reactants
Identifying Limiting Reagents:
1. Convert all given values into moles
2. Divide each mole value by the coefficient
3. The smallest number identifies the LR

18. Stoichiometry - Ch. 3 10. Consider the following unbalanced reaction:
P4H10 (s) + H2O (l)  H3PO4 (s)
a. How many grams of H3PO4 can be produced from a mixture of 10g of each reactant?
b. How many grams of excess reactant remains (assume reaction goes to completion)?
c. What is the percent yield if 22.8 g of H3PO4 is produced?

19. Percent yield = (Actual yield)x(100) (Theoretical yield)
Stoichiometry - Ch. 3
Percent yield = (Actual yield)x(100) (Theoretical yield)

20. Stoichiometry - Ch. 3
11. How many grams of fluorine are required if you want to produce 83 g of PF3 if the reaction has 63.2% yield?
P F2  PF (unbalanced)

21. Stoichiometry - Ch. 3
12. A g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield g of CO2 and g of H2O. What is the empirical formula of the compound?

22. Stoichiometry - Ch. 3
You have completed ch. 3

23. Ch. 3 – Answer Key
1. For a new element, 67.16% is an isotope with mass amu, 2.76% is an isotope with mass amu and 30.08% is an isotope with mass amu. Calculate the average atomic mass for this new element?
Average atomic mass =
(0.6716)(280.8 amu) + (0.0276)(283.7 amu) + (0.3008)(284.8 amu) = amu
2. For which of the following compounds does 1.00 g represent 3.32 × 10-2 mol?
Molar mass is a useful value for identification
Molar mass = (1.00g)/ (3.32 × 10-2 mol) = 30.1g/mol => C2H6

24. Ch. 3 – Answer Key
3. If a sample of diatomic element weighs g and contains 4.162x1024 atoms. Identify the element. Diatomic tells us the formula for the element is X2. To get the molar mass you need the grams (given) and the moles (not given).
4.162x1024 atoms 1molecule X2 1mole X2 = mole X2
Molar mass = 131.3g/3.456 mole = 38.0g/mol => F2
2 atoms
6.022x1023
molecules

25. Ch. 3 – Answer Key 4. If you have 0.63 mg of H2SO4
a. How many H2SO4 molecules are in your sample?
The molar mass of H2SO4 = 2(1.01) (16.0) = g/mol
0.63 mg g mole x1023 molecules = 3.9x1018 molecules
b. How many oxygen atoms are in your sample?
3.9x1018 molecules 4 oxygen atoms = 1.5x10 O atoms
1000 mg
98.08 g
H2SO4
1 mole
1 H2SO4 molecule

26. Ch. 3 – Answer Key
5. An alkali metal oxide contains 83.01% metal by mass. Determine the identity of the metal. How many grams of oxygen are in a 25.0 g sample of the metal oxide? Since an alkali metal has a charge of 1+ the chemical formula for the metal oxide is M2O If the molar mass of the metal is denoted by x ⇒ = ((2x)/(2x + 16))(100) x = 39.1g/mol => K Potassium is the unknown metal

27. Ch. 3 – Answer Key
6. Compound X2Y is 60% X by mass. Calculate the percent Y by mass of the compound XY3?
If you have 100 g of X2Y there would be 60g of X and 40g of Y
For XY3 since it has only one X atom you can think of that as ½(60g) or 30 g of X and since there’s three Y atoms you can think of that as 3(40g) or 120g of Y. So in XY3 there’s 30g of X for every 120g of Y – so %Y = (120g/150g)100 = 80%

28. Ch. 3 – Answer Key
7. Tryptophan is 64.7% carbon, 5.9% hydrogen, 13.7% nitrogen and 15.7% oxygen. What is the empirical formula for tryptophan?
First we need to get a moles for the molar ratio
If we have 100 g sample of typtophan ⇒
64.7 g C/12.01 g/mol = mol C
5.9 g H/1.008 g/mol = mol H
13.7 g N/14.01 g/mol = mol N
15.7 g O/16 g/mol = mol O continue to next slide...

29. Ch. 3 – Answer Key
7. …continued Divide each by the smallest mole value to simplify the ratio ⇒ mol C/0.978 = 5.5 mol C mol H/0.978 = 6 mol H mol N/0.978 = 1 mol N mol O/0.978 = 1 mol O Double each to get whole numbers ⇒ 11 mol C:12 mol H :2 mol N:2 mol O Empirical Formula ⇒ C11H12N2O2

30. Ch. 3 – Answer Key
8. The empirical formula for xylene is C4H5 and xylene has a molar mass of g/mol. Determine the molecular formula for xylene.
First determine the molar mass of the empirical formula ⇒
4(12.01)g/mol + 5(1.008)g/mol = g/mol
Divide molar mass of empirical formula into molar mass of the compound ⇒
(106.16g/mol)/(53.08g/mol) = 2
Multiply the empirical formula by 2⇒
Molecular formula ⇒ C8H10

31. Ch. 3 – Answer Key 9. Consider the following unbalanced reaction:
4 NH O2  4 NO H2O
a. How many moles of oxygen gas are required to make moles of nitrogen dioxide?
12.8 mol NO2 7 mol O2 = 22.4 mol O2
b. How many grams of water can be produced from 9.64 g of ammonia?
9.64 g NH3 1mol NH3 6 mol H2O g H2O = 15.3 g H2O
4 mol NO2
17.04 g NH3
4 mol NH3
1 mol H2O
Continue to next slide…

32. Ch. 3 – Answer Key convert to moles then divide by molar coefficient
9. …continued
c. Identify the limiting reagent if 3 moles of ammonia is combined with 5 moles of oxygen
divide each mole by the molar coefficient and look for smaller value
(3 mol NH3)/4 = 0.75 vs. (5 mol O2)/7 = 0.71 ⇒ O2 is the LR
d. Identify the limiting reagent if 10. g of ammonia is combined with 28 g of oxygen
convert to moles then divide by molar coefficient
(10 g NH3)/( g/mol) = mol NH3/4 = vs.
(28 g O2)/(32 g/mol) = mol O2/7 = ⇒ O2 is the LR

33. Ch. 3 – Answer Key 10. Consider the following unbalanced reaction:
P4H10 (s) + H2O (l)  H3PO4 (s)
a. How many grams of H3PO4 can be produced from a mixture of 10g of each reactant?
First ⇒ balance the reaction ⇒ P4H10 (s) + 6 H2O (l)  4 H3PO4 (s)
Second ⇒ Identify the limiting reagent
10 𝑔 P4H 𝑔/𝑚𝑜𝑙 = mol P4H10 vs 𝑔 H2O 𝑔/𝑚𝑜𝑙 = mol H2O
Limiting reagent
Third ⇒ use LR to find the unknown
mol P4H mol H3PO4 1 mol P4H g H3PO4 1 mol H3PO4 = g H3PO4

34. Ch. 3 – Answer Key 10. continued…
b. How many grams of excess reactant remains (assume reaction goes to completion)?
Limiting reagent will tell you the maximum consumption of the other reactants
mol P4H mol H2O 1 mol P4H g H2O 1 mol H2O = 8.1g of H2O is the maximum amount of H2O that can be consumed
Since there was 10 g of H2O initially then there should be 10g – 8.1g or 1.9g of H2O leftover
c. What is the percent yield if 22.8 g of H3PO4 is produced?
Percent yield = (Actual yield)x(100) (Theoretical yield) = (22.8 g H3PO4)x(100) (29.24 g H3PO4) = 78 %

35. Ch. 3 – Answer Key
11. How many grams of fluorine are needed to produce 83 g of phosphorus trifluoride, if the reaction has 63.2% yield?
P F2  PF (unbalanced)
First ⇒ balance reaction ⇒ P F2  4 PF3
Determine theoretical yield from the actual and the % yield ⇒
63.2% = (83 g/Theo)x100 ⇒ Theoretical yield = g PF3
Use theoretical yield to determine grams of F2
131.3 g PF3 1 mol PF mol F g F2 = 85.1 g F2
87.97 g PF3
4 mol PF3
1 mol F2

36. Ch. 3 – Answer Key CxHyOz + O2  CO2 + H2O ?
12. A g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield g of CO2 and g of H2O. What is the empirical formula of the compound?
CxHyOz + O2  CO2 + H2O
All of the carbon in the compound
will end up in the CO2 and all of the
hydrogen will end up in the water
the oxygen is unpredictable so we need
determine how much C and H there
is in our compound ?
…continue to next slide

37. Ch. 3 – Answer Key 12. …continued C ⇒ 0.8635 g CO2 12.01 g C
H ⇒ g H2O
O ⇒ g cmpd – g C – g H = g O
…continue to next slide
12.01 g C
= g C
44.01 g CO2
2.016 g H
= g H
g H2O

38. Ch. 3 – Answer Key
12. …continued Covert each to moles and divide by smallest value ⇒ ( g C)/(12.01 g/mol) = mol C/ = 1.5 mol C ( g H)/(1.008 g/mol) = mol H/ = 1.5 mol H ( g O)/(16 g/mol) = mol O/ = 1 mol O Double each value 3 mol C:3 mol H:2 mol O Empirical formula ⇒ C3H3O2