Review Sheet – Chemistry, CCP –Moles
Name: ______Date:______Period:___
This test covers materials in the textbook in Chapters 10 (pp. 311-339). You are always responsible for the topics on the previous tests.
. Test Expectations
The student should be able:
· Correctly use the terminology of mass and moles.
· Convert between moles, particles, mass, and volume of a gas (at STP).
· Calculate the percent composition of a compound.
· Determine the empirical and/or molecular formula of a compound from given data.
1. Define each of the following terms:
(a) Formula Unit - One formula weight of an ionic compound.
(b) Molar Mass – Mass in grams of 1 mole of substance.
(c) Avogadro’s number - The number of particles in one mole, equal to 6.02214199 × 1023 mol.
(d) Molar Volume – The volume occupied by one mole of a gas under standard conditions (22.4 L at STP)
(e) Mole - SI unit for amount of substance, defined as the number of atoms in exactly 12 g of carbon-12.
(f) Empirical Formula – Simplest whole number ratio of atoms in a compound.
(g) Molecular Formula – Actual whole number ratio of atoms in a compound.
(h) Atomic Mass – Weighted average of the masses of the isotopes of an element.
(i) STP – Standard condition for gases, 0°C and 1 atm.
(j) Formula Mass – Sum of the all the atomic masses of the atoms in a compound (in amu).
(k) Carbon 12 - Standard for atomic mass and is used to define the mole.
(l) Percent Composition - The percent of each element present in a compound by mass.
(m) Elemental Analysis – Analytical method that determines the percent composition (by mass) of a compound.
(n) Molecule - The smallest particle of a covalent compound that retains the chemical properties of the element or compound.
2. Find the molar mass of the following compounds:
A. Ba(OH)2 1 x Ba 137.33 = 137.33
2 x O 16.00 = 32.00
2 x H 1.01 = 2.02
171.35 g/mole
B. C9H20 9 x C 12.01 = 108.09
20 x H 1.01 = 20.20
128.29 g/mole
3. An accident in a chemistry laboratory releases 7.4 moles of the war gas phosgene (COCl2). How many grams of phosgene were released?
1-step problem
4. A similar accident the previous year released 9.07 x 1024 molecules of phosgene. How much volume did this first accident release? (Assuming this happens at STP…)
2-step problem
5. What is the percent composition by mass of phosgene (COCl2)?
1 x C 12.01 = 12.01
1 x O 16.00 = 16.00
2 x Cl 35.45 = 70.90
= 98.91
6. Mustard gas (C4H8Cl2S) is a poisonous gas that was used in World War I and banned afterward. It causes general destruction of tissues, resulting in the formation of large water blisters. There is no effective antidote. Calculate the percent composition by mass of the elements in mustard gas.
4 x C 12.01 = 48.04
8 x H 1.01 = 8.08
2 x Cl 35.45 = 70.90
1 x S 32.07 = 32.07
159.09 g/mole
7. Ascorbic acid (vitamin C) cures scurvy and may help to prevent the common cold. It is composed of 40.92 % carbon, 4.58 percent hydrogen, and 54.50 percent oxygen by mass. Determine its empirical formula.
STEP 1: Assuming 100 g of ascorbic acid:
STEP 2 & 3:
STEP 4: Multiply by 3 in order to get 1.33 to be a whole number.
3:4:3
C3H4O3
8. The molar mass of caffeine is 194.19 g/mole. Is the molecular formula of caffeine C4H5N2O or C8H10N4O2?
C4H5N2O Molar Mass = 4 x 12 + 5 x 1 + 2 x 14 + 16 = 97 g/mole
The answer is C8H10N4O2, then.
9. The empirical formula of a compound is NaO, and the approximate molar mass is 76 g. What is the molecular formula?
NaO Molar Mass = 23.0 + 16.0 = 39.0 g/mole
The answer then is Na2O2.
10. A compound containing boron and hydrogen has the composition 6.444 g of B and 1.803 g of H. The molar mass of this compound is 30 g. What is its molecular formula?
STEP 1:
STEP 2 & 3:
STEP 4: BH3 is the empirical formula.
STEP 5:
B2H6 is the molecular formula.
11. Allicin is the compound responsible for the characteristic smell of garlic. It contains carbon, hydrogen, sulfur, and oxygen. An analysis of the compound gives the following percent composition by mass: carbon, 44.4%, hydrogen, 6.21%; sulfur 39.5%. What is its molecular formula given that its molar mass is about 162 g/mole.
Let’s try a slightly different method for this problem. Let’s assume a mass equal to the molar mass 162 g/mole. This will make more work for us up front, but simplify the later steps dramatically.
Looking at these values, the actual ratio is clear. The molecular formula is C6H10S2O. To find the empirical formula, the subscripts need to be reduced to its simplest form. In this case, the empirical formula will be the same. This method works for any molecular formula problem (you need the elements molar mass).