Chemical Composition Review Mole Calculations Percent Composition Copyright Cengage Learning. All rights reserved. 8 1
QUESTION Suppose you work in a hardware store and a customer wants to purchase 500 nails. If the average mass of a nail is 20 grams, how many kilograms of nails would you weigh out? 1. 10 kg 2. 10 3 kg 3. 10 2 kg 4. 25 kg Copyright Cengage Learning. All rights reserved. 8 2
ANSWER Choice #1 is the correct answer. This problem illustrates the concept of counting by weighing. Rather than laboriously counting out 500 nails we use the average mass of a nail to compute the mass of 500 nails and simply weigh out the correct mass. 20 g 1 kg 500 nails 10 kg 1 nail 1000 g Copyright Cengage Learning. All rights reserved. 8 3
QUESTION A pile of marbles weigh 394.80 g. 10 marbles weigh 37.60 g. How many marbles are in the pile? 1. 11 2. 40 3. 105 4. 223 Copyright Cengage Learning. All rights reserved. 8 4
ANSWER Choice #3 is the correct answer. Avg. Mass of 1 Marble = 37.60 g 10 marbles = 3.76 g / marble 394.80 g = 105 marbles 3.76 g Copyright Cengage Learning. All rights reserved. 8 5
QUESTION Atomic masses on the periodic table are not reported as whole numbers. What is the chief reason for this? 1. Atoms are so small that even modern techniques have a limited accuracy, which keeps the value from being a whole number. 2. Atomic mass scales are based on hydrogen 1.01, and since this atom does not have a whole number mass the comparisons will not be able to have whole number masses also. 3. Isotopic abundance is averaged to provide the mass of atoms. 4. Atomic mass is based on a variable scale. Copyright Cengage Learning. All rights reserved. 8 6
ANSWER Choice #3 properly explains that the masses of elements are based on the weighted average of the isotopes for each element. Copyright Cengage Learning. All rights reserved. 8 7
QUESTION Since it is extremely difficult to see atoms without elaborate equipment, chemists use the atomic mass scale to determine the number of atoms in a sample. For example, calculate the number of aluminum atoms in a piece of aluminum foil that weighs 2.58 10 8 amu. 1. 9.56 x 10 6 2. 2.58 x 10 8 3. 2.58 x 10 6 4. 1.33 x 10 20 Copyright Cengage Learning. All rights reserved. 8 8
ANSWER Choice #1 provides the correct answer. 8 1 atom Al 6 2.58 10 amu 9.56 10 atoms Al 26.98 amu Copyright Cengage Learning. All rights reserved. 8 9
QUESTION Diamond is an allotrope of carbon. If 1.0 carat has a mass of 0.20 grams, how many carbon atoms are in a 3.0 carat diamond? 1. 6.0 x 10 15 2. 7.2 x 10 10 3. 3.0 x 10 26 4. 3.0 x 10 22 Copyright Cengage Learning. All rights reserved. 8 10
ANSWER Choice #4 correctly gives the atomic make up of the 3.0 carat diamond. 0.20 g C 1 amu C 1 atom C 3.0 c 24 1.0 c 1.66 10 g C 12.01 amu C 22 3.0 10 atoms C Copyright Cengage Learning. All rights reserved. 8 11
QUESTION Calculate the number of iron atoms in a 4.48 mole sample of iron. 1. 7.44 10 24 2. 1.34 10 23 3. 12.5 4. 2.70 10 24 Copyright Cengage Learning. All rights reserved. 8 12
ANSWER Choice #4 is the correct answer. 23 6.022 10 atoms Fe 24 4.48 mol Fe 2.70 10 atoms Fe 1 mol Fe Copyright Cengage Learning. All rights reserved. 8 13
QUESTION The metal gallium (Ga) has most of the properties you would expect in a metal, but it is so soft it will melt in your hand. If you had 20.0 grams of solid gallium in your palm and watched it turn into liquid, how many moles of liquid gallium would you soon have? 1. 20.0 moles 2. 0.287 moles 3. 4.32 moles 4. 3.6 10 23 moles Copyright Cengage Learning. All rights reserved. 8 14
ANSWER Choice #2 represents the mole equivalent for 20.0 grams of gallium. 1 mole Ga 20.0 g Ga 0.287 moles Ga 69.7 g Ga Copyright Cengage Learning. All rights reserved. 8 15
QUESTION Which of the following is closest to the average mass of one atom of copper? 1. 63.55 g 2. 58.93 g 3. 65.38 g 4. 1.055 x 10 22 g Copyright Cengage Learning. All rights reserved. 8 16
ANSWER Choice #4 is the correct answer. The mass of one atom of copper is going to be extremely small, so only Choice #4 makes sense. The calculated solution is: 1 mol Cu 63.55 g Cu 22 1 Cu atom = 1.055 10 g Cu 23 6.022 10 Cu atoms 1 mol Cu Copyright Cengage Learning. All rights reserved. 8 17
QUESTION A sample of 26.98 grams of Al has the same number of atoms as grams of Au. 1. 26.98 g 2. 13.49 g 3. 197.0 g 4. 256.5 g Copyright Cengage Learning. All rights reserved. 8 18
ANSWER Choice #3 gives the mass of 1.0 mole of gold. The 26.98 g of Al represents 1.0 mole of Al. Thus both samples contain Avogadro s number, 6.022 10 23 atoms. Copyright Cengage Learning. All rights reserved. 8 19
QUESTION Which silver ring would be worth more? 1. A ring that weighs 10.0 grams. 2. A ring that consists of 1.0 10 22 atoms of Ag. Copyright Cengage Learning. All rights reserved. 8 20
ANSWER Choice #1 is the correct answer. The ring that weighs 10.0 g would be worth more. We can tell by using dimensional analysis to figure out how much the 1.0 10 22 atoms of Ag would weigh: 22 1 mole Ag 107.9 g Ag 1.0 10 atoms Ag 1.8 g Ag 23 6.022 x 10 atoms Ag 1 mole Ag Copyright Cengage Learning. All rights reserved. 8 21
QUESTION Which of the following 100.0 g samples contains the greatest number of atoms? 1. Magnesium 2. Zinc 3. Silver Copyright Cengage Learning. All rights reserved. 8 22
ANSWER Choice #1 is the correct answer. Magnesium has the smallest average atomic mass (24.31 amu). Since magnesium is lighter than zinc and silver, it will take more atoms to reach 100.0 g. Copyright Cengage Learning. All rights reserved. 8 23
QUESTION You have two beakers on your lab table. Beaker #1 contains 32.07 g of sulfur (S) and Beaker #2 contains 74.92 g of arsenic (As). Which beaker contains the greatest number of atoms? Choose the best answer. 1. Beaker #1 because one sulfur atom weighs less than one arsenic atom so you need more of the sulfur atoms to fill the beaker. 2. Beaker #2 because arsenic has a greater mass than sulfur. 3. Beaker #2 because it contains more moles of arsenic atoms versus the number of moles of sulfur atoms in Beaker #1. 4. Beakers #1 and #2 contain the same number of atoms because there is one mole in each. Copyright Cengage Learning. All rights reserved. 8 24
ANSWER Choice #4 is the correct answer. They each contain one mole of atoms, which is equivalent to Avogadro s number. Copyright Cengage Learning. All rights reserved. 8 25
QUESTION One mole of carbon is equivalent to 1. 12.01 atomic mass units. 2. the same mass of atoms that are contained in one mole of calcium. 3. the same number of atoms that are contained in one mole of calcium. 4. 6.022 10 23 grams of carbon. Copyright Cengage Learning. All rights reserved. 8 26
ANSWER Choice #3 is the correct answer. Copyright Cengage Learning. All rights reserved. 8 27
QUESTION What is the molar mass of nickel(ii) carbonate? 1. 118.7 g/mol 2. 134.7 g/mol 3. 178.71 g/mol 4. 296.09 g/mol Copyright Cengage Learning. All rights reserved. 8 28
ANSWER Choice #1 is the correct answer. The formula for nickel(ii) carbonate is NiCO 3. Therefore the molar mass is: 58.69 + 12.01 + 3(16.00) = 118.7 g/mol. Copyright Cengage Learning. All rights reserved. 8 29
QUESTION The molar mass of ammonium phosphate is: 1. 131.05 g/mol 2. 144.06 g/mol 3. 149.10 g/mol 4. 302.95 g/mol Copyright Cengage Learning. All rights reserved. 8 30
ANSWER Choice #3 is the correct answer. The formula for ammonium phosphate is (NH 4 ) 3 PO 4. Therefore the molar mass is: 3(14.01) + 12(1.008) + 30.97 + 4(16.00) = 149.10 g/mol. Copyright Cengage Learning. All rights reserved. 8 31
QUESTION Consider equal mole samples of N 2 O, Al(NO 3 ) 3, and KCN. Rank these from least to most number of nitrogen atoms in each sample. 1. KCN, Al(NO 3 ) 3, N 2 O 2. Al(NO 3 ) 3, N 2 O, KCN 3. KCN, N 2 O, Al(NO 3 ) 3 4. Al(NO 3 ) 3, KCN, N 2 O Copyright Cengage Learning. All rights reserved. 8 32
ANSWER Choice #3 is the correct answer. Assume you have one mole of each sample. This means you have 1 mole of N in KCN, 2 moles of N in N 2 O and 3 moles of N in Al(NO 3 ) 3. Moles are directly related through Avogadro s number. Copyright Cengage Learning. All rights reserved. 8 33
QUESTION You have 27.0 g of an unknown substance A and 75.0 g of chlorine gas. Substance A contains 1.5 times as many molecules as the chlorine gas. What is the identity of A? 1. HClO 2. NO 2 3. BF 3 4. NH 3 Copyright Cengage Learning. All rights reserved. 8 34
ANSWER Choice #4 is the correct answer. To identify A, you need to know the molar mass (# grams/# moles). The number of grams of A is already given. To find the moles of A: 1 mol Cl 1.5 mol A 2 75.0 g Cl = 1.6 mol A 2 70.9 g Cl 1 mol Cl 2 2 Therefore the molar mass of A is 17 g/mol (27.0 g/1.6 mol), which is NH 3. Copyright Cengage Learning. All rights reserved. 8 35
QUESTION How many grams of fluorine are contained in one molecule of boron trifluoride? 1. 3.155 10 23 g 2. 9.465 10 23 g 3. 6.022 10 23 g 4. 3.433 10 25 g Copyright Cengage Learning. All rights reserved. 8 36
ANSWER Choice #2 is the correct answer. 1 molecule BF 3 23 = 9.465 10 g F 3 atoms F 1 mol F 19.00 g F 23 1 molecule BF 6.022 10 atoms F 1 mol F 3 Copyright Cengage Learning. All rights reserved. 8 37
QUESTION Morphine, derived from opium plants, has the potential for use and abuse. It s formula is C 17 H 19 NO 3. What percent, by mass, is the carbon in this compound? 1. 12.0 % 2. 54.8 % 3. 67.9 % 4. 71.6 % Copyright Cengage Learning. All rights reserved. 8 38
ANSWER Choice #4 is correct. First determine the molar mass of the compound, then divide that into the total mass of carbon present. Finally multiply that by 100 to convert from a fraction into a percentage. 17 12.01 17 12.01 + 19 1.008 + 14.01 + 3 16.00 = 0.716 100 = 71.6% C Copyright Cengage Learning. All rights reserved. 8 39
QUESTION What is the mass percent of potassium in potassium nitride? 1. 89.33% 2. 48.19% 3. 38.67% 4. 29.78% Copyright Cengage Learning. All rights reserved. 8 40
ANSWER Choice #1 is the correct answer. The formula for potassium nitride is K 3 N. 3 39.10 3 39.10 + 14.01 = 0.8933 100 = 89.33% K Copyright Cengage Learning. All rights reserved. 8 41
QUESTION What is the percent mass of phosphorus in calcium phosphide? 1. 19.97% 2. 34.00% 3. 43.59% 4. 53.68% Copyright Cengage Learning. All rights reserved. 8 42
ANSWER Choice #2 is the correct answer. The formula for calcium phosphide is Ca 3 P 2. 2 30.97 3 40.08 + 2 30.97 = 0.3400 100 = 34.00% P Copyright Cengage Learning. All rights reserved. 8 43
QUESTION A gas consisting of 50.0% sulfur and 50.0% oxygen by mass has what empirical formula? 1. SO 2. SO 2 3. S 2 O 4. SO 4 Copyright Cengage Learning. All rights reserved. 8 44
ANSWER Choice #2 is the smallest whole number ratio of the atoms that make up this compound. When presented with % composition data, we start out by assuming we have 100 g of that compound so that the % s are masses in grams. We then convert the masses into moles using atomic masses: 1 mole S 50.0 g S 1.56 mole S 32.06 g S 1 mole O 50.0 g O 3.13 mole O 16.00 g O We then divide by the smallest # of moles: 3.13/1.56 = 2 Therefore # moles O/# moles S = 2/1 SO 2 Copyright Cengage Learning. All rights reserved. 8 45
QUESTION A hydrocarbon (binary compound containing carbon and hydrogen) has a percent composition of 92% carbon and 8% hydrogen and a molar mass of about 26 g/mol. Determine the molecular formula of this compound. 1. CH 4 2. C 2 H 4 3. C 2 H 2 4. CH 2 Copyright Cengage Learning. All rights reserved. 8 46
ANSWER Choice #3 is the only formula listed with a molar mass of 26 g/mol. To see how the formula is worked out, we convert the % s to grams and divide by atomic masses: 1 mole C 92 g C 7.7 mole C 12.01 g C 1 mole H 8 g H x 8 mole H 1.008 g H 7.7 rounds off to 8 so C/H is 1/1 in the compound. The empirical formula is CH, but this has a molecular weight of 13. 26/13 = 2, so we multiply the formula CH by 2 to give C 2 H 2. Copyright Cengage Learning. All rights reserved. 8 47
QUESTION Vinegar contains carbon, hydrogen, and oxygen with percent masses of 40.01% C and 6.70% H, respectively. If the molar mass of vinegar is about 60 g/mol, what is its molecular formula? 1. CH 2 O 2. C 3 H 8 O 3. C 2 H 20 O 4. C 2 H 4 O 2 Copyright Cengage Learning. All rights reserved. 8 48
ANSWER Choice #4 is the correct answer. The empirical formula is CH 2 O with a molar mass of about 30 g/mol. 60/30 = 2, so we multiply the formula CH 2 O by 2 to get C 2 H 4 O 2. Copyright Cengage Learning. All rights reserved. 8 49
QUESTION A compound consists of silicon and 8.23% by mass of hydrogen. The molar mass of the compound is approximately 122 g/mol. What is the molecular formula of this compound? 1. Si 4 H 10 2. SiH 2 3. SiH 3 4. Si 2 H 5 Copyright Cengage Learning. All rights reserved. 8 50
ANSWER Choice #1 is the correct answer. The empirical formula is Si 2 H 5 with a molar mass of about 61 g/mol. 122/61 = 2, so we multiply the formula Si 2 H 5 by 2 to get Si 4 H 10. Copyright Cengage Learning. All rights reserved. 8 51