1. Chapter 10 Answers

2. Mole-Particle Conversions
#1) How many formula units are in 0.89 moles CaCl2?
1 mole = 6.02x1023 fu
#2) A bottle of PbSO4 contains 4.25 x 1024 formula units of the compound. How many moles?
1 mole = 6.02x1023 fu

3. Mole-Particle Conversions
#3) How many molecules are in 3.5 mole CH4?
1 mole = 6.02x1023 molecules
#4) A tube with 3.68 moles of Neon contains how many atoms?
1 mole = 6.02x1023 atoms

4. Mole-Particle Conversions
#5) 1 mole = 6.02x1023 molecules
#6) 1 mole = 6.02x1023 ions
#7) 1 mole = 6.02x1023 atoms

5. Mole-Mass Conversions
#1) How many grams are in 0.75 moles of MgF2?
1 mole = molar mass (g)
1 mole = 62.3 g
#2) A student measures g Ca(NO3)2. How many moles?
1 mole = molar mass (g)
1 mole = g

6. Mole-Mass Conversions
#3) How many molecules are in 14.0 moles of H2O?
1 mole = 6.02x1023 molecules
#4) A tube that contains 3.68 moles of Neon would be what mass?
1 mole = molar mass (g)
1 mole = 20.2 g

7. Mole-Mass Conversions
#5) 1 mole = molar mass (g)
1 mole = 63.0 g
#6) 1 mole = 6.02 x 1023 fu
#7) 1 mole = molar mass (g)
1 mole = 63.5 g

8. 1 step review 1 mole = molar mass (g) 1 mole = 181.0 g
#1) You have 23 moles of Tantalum (Ta). How many grams is this?
1 mole = molar mass (g)
1 mole = g
#2) The label says that the bottle holds 500 grams of Al2(CO3)3. How many moles are there?
1 mole = molar mass (g)
1 mole = g

9. 1 step review 1 mole = 22.4 L 1 mole = molar mass (g) 1 mole = 107.9 g
#3) How many moles of argon atoms are present in 11.2 L of argon gas at STP?
1 mole = 22.4 L
#4) Your silver watchband masses out at 326 g. How many moles do you have?
1 mole = molar mass (g)
1 mole = g

10. 1 step review 1 mole = 6.02 x 1023 molecules 1 mole = 22.4 L
#5) How many molecules are in moles of N2O5?
1 mole = 6.02 x 1023 molecules
#6) Determine the volume, in liters, occupied by moles of a gas at STP.
1 mole = 22.4 L

11. 1 step review #7) 1 mole = 342.0 g #8) 1 mole = 6.02x1023 atoms
#9) 1 mole = 22.4 L

12. 1 step review #10) 1 mole = 6.02x1023 fu #11) 1 mole = 60.0 g
#12) 1 mole = 22.4 L

13. Two Step Conversions 1 mole = 6.02 x 1023 molecules 1 mole = 30.0 g
#1) If you burn 6.10 x 1024 molecules of C2H6, what mass of ethane did you burn?
1 mole = 6.02 x 1023 molecules
1 mole = 30.0 g
#2) A chemical reaction produces 1.38 L of HBr gas. How many grams?
1 mole = 22.4 L
1 mole = 80.9 g

14. Two Step Conversions 1 mole = 40.0g 1 mole = 6.02 x 1023 fu
#3) How many formula units are in 13.5 g of NaOH?
1 mole = 40.0g
1 mole = 6.02 x 1023 fu
#4) How many atoms are in 65 L of Argon gas at STP?
1 mole = 22.4 L
1 mole = 6.02 x 1023 atoms

15. Two Step Conversions 1 mole = 6.02 x 1023 molecules 1 mole = 22.4 L
#5) How much volume would 8.4 x 1021 molecules of H2S gas occupy at STP?
1 mole = 6.02 x 1023 molecules
1 mole = 22.4 L
#6) If I measure out 234 g of sugar (C6H22O11), how many molecules do I have?
1 mole = g
1 mole = 6.02 x 1023 molecules

16. Two Step Conversions

17. All Mole Conversions 1 mole = 12.0g 1 mole = 6.02 x 1023 molecules
#1) How many grams of carbon are in 3.5 moles?
1 mole = 12.0g
#2) If I have 1.6 x 1024 molecules of FeCl3, how many grams do I have?
1 mole = 6.02 x 1023 molecules
1 mole = g

18. All Mole Conversions 1 mole = 22.4 L 1 mole = 6.02 x 1023 molecules
#3) How many grams are in a 5 L tank of NO2 (at STP)?
1 mole = 22.4 L
1 mole = 6.02 x 1023 molecules
#4) How many moles of Zinc are in 1.65 x 1023 atoms?
1 mole = 6.02 x 1023 atoms

19. All Mole Conversions 1 mole = 22.4 L 1 mole = 4.0 g
#5) There are about 700,000 L of Helium in a Macy’s Thanksgiving Day parade balloon. How many grams of Helium would this be (at STP)?
1 mole = 22.4 L
1 mole = 4.0 g
#6) If I have 0.8 moles of sugar in my tea, how many molecules of sugar would I have?
1 mole = 6.02 x 1023 molecules

20. All Mole Conversions #7) 1 mole = 6.02 x 1023 atoms 1 mole = 22.4 L
#8) 1 mole = g
1 mole = 6.02 x 1023 atoms
#9) 1 mole = 44.0 g
1 mole = 6.02 x 1023 molecules

21. All Mole Conversions

22. Percent Composition #1) NaOH
Na – 23.0 x 1 = /40.0 x 100 = % Na
O – 16.0 x 1 = /40.0 x 100 = % O
H – x 1 = /40.0 x 100 = % H
40.0g
#2) (NH4)2SO4
N – 14.0 x 2 = /132.1 x 100 = 21.2% N
H – x 8 = /132.1 x 100 = % H
S – x 1 = /132.1 x 100 = 24.3% S
O – 16.0 x 4 = /132.1 x 100 = % O
132.1g

23. Percent Composition Al – 1.30/2.45 x 100 = 53.1% Al
#3) A sample of 2.45 g of aluminum oxide decomposes into 1.30 g of aluminum and 1.15 g of oxygen. What is the percent composition of the compound?
Al – 1.30/2.45 x 100 = % Al
O – 1.15/2.45 x 100 = % O
#4) Find the percent composition of a compound that contains 1.94 g of carbon, g of hydrogen and 2.58 g of sulfur.
whole = = 5.00
C – 1.94/5.00 x 100 = 38.8% C
H – 0.48/5.00 x 100 = % H
S – 2.58/5.00 x 100 = % S

24. Percent Composition C – 12.0 x 6 = 72.0 72.0/180.0 x 100 = 40.0 % C
#5) Determine the mass of Carbon in 120 g C6H12O6
C – 12.0 x 6 = /180.0 x 100 = % C
H – 1.0 x 12 = 12.0
O –16.0 x 6 = 96.0
180.0g
40.0 % of 120g
0.400 x 120 = 48g
#6) What is the mass of magnesium in 23.2 g MgCl2?
Mg – 24.3 x 1 = /95.3 x 100 = 25.5% Mg
Cl – x 2 = 71.0
95.3g
25.5 % of 23.2g
0.255 x 23.2 = 5.92g

25. Percent Comp. Back 1 mole = 148.3 g #1) Aluminum Chromate = Al2(CrO4)3
#2) How many moles are in 23 g of magnesium nitrate? = Mg(NO3)2
1 mole = g

26. Percent Comp. Back 1 mole = 142.1 g 1 mole = 6.02x1023 molecules
#3) How many grams are in 5.6 x 1022 molecules of sodium sulfate?
Na2SO4
1 mole = g
#4) If you have 3.5 moles of CaBr2, how many molecules do you have?
1 mole = 6.02x1023 molecules

27. Percent Comp. Back
#5) A balloon filled with 7.3 x 1025 atoms of helium at STP would take up how many liters? How many grams of helium would this be?
#6) If your lungs could hold about 6 liters of oxygen (O2) at STP, how many grams of oxygen would that be?

28. Empirical Formulas
#1) What is the empirical formula of a compound that contains % Na; % S; and % O?

29. Empirical Formulas
#2)
#3)

30. Empirical Formulas
#4)
#5) g – g = g

31. Empirical Formulas Back

32. Empirical Formulas Back
#3)
C – 12.0 x 2 = /45.0 x 100 = % C
H – 1.0 x 5 = /45.0 x 100 = % H
O –16.0 x 1 = /45.0 x 100 = % O
45.0g
#4)
S – 32.1 x 1 = /146.1 x 100 = % S
F –19.0 x 6 =
146.1g
22.0 % of 230g
0.220 x 230 = 50.6 g

33. Molecular Formulas
#1)
#2)

34. Molecular Formulas
#3)

35. Ch 10 Problem Review #1) What is the atomic mass of Zn?
65.4 amu *Note units
#2) What is the molar mass of Al2(SO4)3?
342.3 g *Note units
#3) Find the mass of moles of SiO2.
#4) How many molecules are found in 3.65 moles of AgNO3?

36. Ch 10 Problem Review Zn – 65.4 x 1 = 65.4 65.4/183.4 x 100 = 35.7% Zn
#5) If a container holds 6.5 L of CO2, what mass of gas is enclosed?
#6) How many grams of mass of C2H6 would 2.37 x 1023 molecules of C2H6 contain?
#7) If you have Zn(C2H3O2)2, what is the percent composition?
Zn – 65.4 x 1 = /183.4 x 100 = 35.7% Zn
C – x 4 = /183.4 x 100 = % C
H – x 6 = /183.4 x 100 = % H
O – 16.0 x 4 = /183.4 x 100 = % O
183.4g

37. Ch 10 Problem Review
#8) Phenylalanine is an essential amino acid whose chemical composition is 65.5% C, 6.67% H, 8.48% N and 19.4% O. What is the empirical formula ?

38. Ch 10 Problem Review
#9) Determine the empirical formula of a compound containing 24.74% potassium, 34.76% manganese, and 40.50% oxygen.
#10) Find the empirical formula for a compound that contains 6.5 g potassium, 5.9 g chlorine and 8.0 g oxygen.

39. Ch 10 Problem Review
11.)

40. Chapter 10 Review What is a mole?
-The number of atoms of an element equal to the number of atoms in exactly 12.0 g of carbon-12.
(That number happens to be 6.02x1023)
Why do we use the mole?
-To simplify very large numbers
Avagadro’s number? 6.02x1023 – It is the number of particles in a mole of any substance.

41. Chapter 10 Review Formula Mass?
-The sum of the atomic masses in a compound (in amu’s)
Molar Mass?
-The mass of one mole of particles (in grams)
Related?
-Molar mass is equal to formula mass (or atomic mass) but in grams.

42. Chapter 10 Review Relationships -mass & moles: 1 mole = molar mass (g)
-molecules/atoms/fu & moles: 1 mole = 6.02x1023 particles
-volume & moles: 1 mole = 22.4 L
Molar volume conditions?
-STP (Standard Temp & Pressure) = 1 atm & 0°C

43. Chapter 10 Review Empirical Formula?
-The simplest (reduced) formula (ratio of atoms)
Molecular Formula?
-Not reduced – so actual number of atoms present
Related?
-molecular formula = (empirical formula)x
(molecular is empirical times some multiplier)

44. Chapter 10 Review
Problems:
1.)
2.)
3.)
4.)

45. Chapter 10 Review Problems:
5.) Na – 23.0 x 2 = /134.0 x 100 = % Na
C – 12.0 x 2 = /134.0 x 100 = % C
O – 16.0 x 4 = /134.0 x 100 = % O
134.0g
6.) Copper(II)oxide = CuO
Cu – 63.5 x 1 = /79.5 x 100 = % Cu
O – 16.0 x 1 = 16.0
79.5g x 325g = 259.7g

46. Chapter 10 Review 7.) 8.)C2HCl C – 12.0 x 2 = 24.0 H – 1.0 x 1 = 1.0
Cl – 35.5 x 1 = 35.5
60.5g

47. Ch 10 Review
9.)

48. Ch 10 Review
10.)

49. Ch 10 Practice Test Multiple Choice: B A C D Problems:
11. Na – 23.0 x 1 = 23.0
H – x 1 = 1.0
C – x 1 = 12.0
O – x 3 = 48.0
84.0 amu
Ba – x 1 = 137.3
O – x 2 =
H – x 2 =
171.3 grams

50. Chapter 10 Review
13.)
14.)
15.)
16.)

51. Chapter 10 Review 17.) Fe – 2.232/3.192 x 100 = 69.9% Fe
O – 0.96/3.192 x 100 = % O
18.)

52. Ch 10 Review
19.)

53. Chapter 10 Review
22.)
23.)
24.) H – x 4 = /60.0 x 100 = % H
C – 12.0 x 2 = /60.0 x 100 = % C
O – 16.0 x 2 = /60.0 x 100 = % O
60.0g