Using atomic masses of C = 12.01 amu, H = 1.008 amu, and O = 16.00 amu, what is the molecular mass of glucose, C6H12O6, to two decimal places?
180.16 amu
✓
29.02 amu
168.06 amu
174.11 amu
Correct Answer
180.16 amu
Multiply each atomic mass by its subscript and add: 6(12.01) + 12(1.008) + 6(16.00) = 72.06 + 12.096 + 96.00 = 180.16 amu. 29.02 amu adds one atom of each element and ignores the subscripts. 168.06 amu leaves out the hydrogen entirely. 174.11 amu counts only 6 hydrogen atoms instead of 12.
Category: Formula Mass
Question 2
Sodium chloride is an ionic compound that does not exist as discrete molecules. The sum of the average atomic masses, in amu, of the ions shown in its formula NaCl is properly called its
Molecular mass
Formula mass
✓
Molar mass
Atomic number
Correct Answer
Formula mass
Formula mass is the general term for the summed atomic masses in any formula, and it is the correct term for ionic compounds, whose formulas describe a formula unit rather than a molecule. Molecular mass applies only to covalent substances made of discrete molecules. Molar mass is the mass of one mole in grams per mole, not a value in amu for one unit. Atomic number counts protons in a single atom and is not a mass.
Category: Formula Mass
Question 3
A sample of water has a mass of 36.04 g. Using a molar mass of 18.02 g/mol for H2O, how many moles of water are in the sample?
0.5000 mol
649.4 mol
2.000 mol
✓
18.02 mol
Correct Answer
2.000 mol
Moles equal mass divided by molar mass: 36.04 g / 18.02 g/mol = 2.000 mol. 0.5000 mol comes from dividing molar mass by mass, the inverse of the correct setup. 649.4 mol comes from multiplying mass by molar mass. 18.02 mol simply repeats the molar mass as if it were the amount.
Category: The Mole
Question 4
Using 6.022 x 10^23 per mole for Avogadro's number, how many molecules are in 0.500 mol of carbon dioxide, CO2?
1.20 x 10^24 molecules
6.02 x 10^23 molecules
9.03 x 10^23 molecules
3.01 x 10^23 molecules
✓
Correct Answer
3.01 x 10^23 molecules
Multiply moles by Avogadro's number: 0.500 mol x 6.022 x 10^23 per mol = 3.01 x 10^23 molecules. 1.20 x 10^24 molecules results from dividing Avogadro's number by 0.500 instead of multiplying. 6.02 x 10^23 molecules is the count for a full mole, not half a mole. 9.03 x 10^23 molecules is actually the number of atoms, since each CO2 molecule has three atoms.
Category: The Mole
Question 5
Using C = 12.01 and O = 16.00, what is the mass percentage of carbon in carbon dioxide, CO2, to two decimal places?
27.29%
✓
42.88%
72.71%
33.33%
Correct Answer
27.29%
The molar mass of CO2 is 12.01 + 2(16.00) = 44.01 g/mol, so carbon is 12.01 / 44.01 x 100% = 27.29% of the mass. 42.88% uses only one oxygen atom in the formula mass. 72.71% is the mass percentage of oxygen, not carbon. 33.33% is the fraction of atoms that are carbon, which is not the same as the mass fraction.
Category: Percent Composition
Question 6
Hydrogen peroxide has the molecular formula H2O2. What is its empirical formula?
H2O
HO
✓
H2O2
HO2
Correct Answer
HO
An empirical formula shows the simplest whole-number ratio of atoms, and dividing both subscripts of H2O2 by 2 gives HO. H2O is water, which has a 2:1 ratio rather than 1:1. H2O2 is the molecular formula and can still be reduced. HO2 has a 1:2 ratio, which does not match the compound.
Category: Empirical Formulas
Question 7
A compound is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Using C = 12.01, H = 1.008, and O = 16.00, what is its empirical formula?
CHO
C2HO
CH2O
✓
CH2O2
Correct Answer
CH2O
In a 100 g sample there are 40.0/12.01 = 3.33 mol C, 6.7/1.008 = 6.65 mol H, and 53.3/16.00 = 3.33 mol O; dividing by 3.33 gives a 1:2:1 ratio, or CH2O. CHO ignores that there are about twice as many moles of hydrogen as carbon. C2HO comes from mixing up the carbon and hydrogen amounts. CH2O2 would require twice as many moles of oxygen as carbon.
Category: Empirical Formulas
Question 8
A hydrocarbon has the empirical formula CH2, whose empirical unit has a mass of 14.03 amu, and a molar mass of 56.1 g/mol. What is its molecular formula?
C2H4
CH2
C8H16
C4H8
✓
Correct Answer
C4H8
Dividing the molar mass by the empirical formula mass gives 56.1 / 14.03 = 4.00, so each subscript is multiplied by 4 to give C4H8. C2H4 has a molar mass of only about 28 g/mol. CH2 is the empirical formula itself, which is too light. C8H16 has a molar mass of about 112 g/mol, twice the measured value.
Category: Empirical Formulas
Question 9
In determining the empirical formula of an iron oxide, the mole ratio of Fe to O works out to 1.00 : 1.50. What is the empirical formula?
Fe2O3
✓
FeO2
FeO
Fe3O2
Correct Answer
Fe2O3
A ratio of 1.00 : 1.50 is not a whole-number ratio, so both numbers are multiplied by 2 to give 2 : 3, which is Fe2O3. FeO2 comes from incorrectly rounding 1.50 up to 2. FeO comes from incorrectly rounding 1.50 down to 1. Fe3O2 reverses the subscripts of the two elements.
Category: Empirical Formulas
Question 10
A solution is made by dissolving 0.50 mol of NaCl in enough water to make 2.0 L of solution. What is its molarity?
1.0 M
0.25 M
✓
4.0 M
0.50 M
Correct Answer
0.25 M
Molarity is moles of solute divided by liters of solution: 0.50 mol / 2.0 L = 0.25 M. 1.0 M comes from multiplying moles by volume. 4.0 M comes from dividing liters by moles, the inverse of the correct setup. 0.50 M ignores the volume and reports only the moles.
Category: Molarity
Question 11
A student dissolves 4.00 g of NaOH (molar mass 40.00 g/mol) in enough water to make 250. mL of solution. What is the molarity of the solution?
0.0160 M
0.100 M
0.400 M
✓
16.0 M
Correct Answer
0.400 M
First convert mass to moles: 4.00 g / 40.00 g/mol = 0.100 mol NaOH. Then divide by the volume in liters: 0.100 mol / 0.250 L = 0.400 M. 0.0160 M divides grams by milliliters and skips both conversions. 0.100 M is the number of moles, not a concentration. 16.0 M divides grams directly by liters without converting to moles.
Category: Molarity
Question 12
How many moles of KCl are present in 250. mL of a 0.200 M KCl solution?
50.0 mol
0.800 mol
1.25 mol
0.0500 mol
✓
Correct Answer
0.0500 mol
Moles of solute equal molarity times volume in liters: 0.200 mol/L x 0.250 L = 0.0500 mol. 50.0 mol multiplies by 250 mL without converting milliliters to liters. 0.800 mol divides molarity by volume. 1.25 mol divides volume by molarity.
Category: Molarity
Question 13
A chemist dilutes 50.0 mL of 6.00 M HCl with water to a final volume of 300. mL. What is the concentration of the diluted solution?
1.00 M
✓
36.0 M
1.20 M
6.00 M
Correct Answer
1.00 M
The moles of solute do not change on dilution, so M1V1 = M2V2 gives M2 = (6.00 M)(50.0 mL) / 300. mL = 1.00 M. 36.0 M puts the volumes in the wrong places, which wrongly makes the solution more concentrated. 1.20 M uses 250 mL, the volume of water added, instead of the final volume. 6.00 M is the original concentration, which must decrease when water is added.
Category: Dilution
Question 14
What volume of 12.0 M HCl stock solution is needed to prepare 500. mL of 3.00 M HCl?
2.00 x 10^3 mL
125 mL
✓
41.7 mL
375 mL
Correct Answer
125 mL
Using M1V1 = M2V2, V1 = (3.00 M)(500. mL) / 12.0 M = 125 mL of stock, which is then diluted to 500. mL. 2.00 x 10^3 mL places the two concentrations the wrong way around. 41.7 mL divides 500 mL by 12.0 M and leaves out the target concentration. 375 mL is roughly the amount of water added, not the volume of stock needed.
Category: Dilution
Question 15
A solution is made by dissolving 15.0 g of sugar in 85.0 g of water. What is the mass percentage of sugar in the solution?
17.6%
85.0%
15.0%
✓
1.50%
Correct Answer
15.0%
Mass percentage is solute mass divided by solution mass times 100%, and the solution mass is 15.0 g + 85.0 g = 100.0 g, so the answer is 15.0%. 17.6% divides by the mass of water alone instead of the whole solution. 85.0% is the mass percentage of water. 1.50% misplaces the decimal point.
Category: Concentration Units
Question 16
A 1.0 kg sample of tap water contains 2.0 mg of fluoride. What is the fluoride concentration in parts per million by mass?
2.0 x 10^3 ppm
2.0 x 10^-3 ppm
2.0 x 10^-6 ppm
2.0 ppm
✓
Correct Answer
2.0 ppm
Parts per million is (mass of solute / mass of solution) x 10^6, so (0.0020 g / 1000 g) x 10^6 = 2.0 ppm. A convenient shortcut for dilute water samples is that 1 mg per kg equals 1 ppm. 2.0 x 10^3 ppm treats the 2.0 mg of fluoride as if it were 2.0 g. 2.0 x 10^-3 ppm is the mass fraction in grams per kilogram, not ppm. 2.0 x 10^-6 ppm is the plain mass fraction, which was never multiplied by 10^6.
Category: Concentration Units
Question 17
A solution is made by mixing 35 mL of ethanol with water to give 250 mL of solution. What is the volume percentage of ethanol, to two significant figures?
14%
✓
16%
35%
1.4%
Correct Answer
14%
Volume percentage is solute volume divided by solution volume times 100%: 35 mL / 250 mL x 100% = 14%. 16% divides by 215 mL, an estimate of the water alone, instead of the total solution volume. 35% reports the ethanol volume as if it were a percentage. 1.4% misplaces the decimal point.
Category: Concentration Units
Question 18
Physiological saline is 0.90% (m/v) NaCl, meaning 0.90 g of NaCl per 100 mL of solution. How many grams of NaCl are in 500. mL of this saline?
0.45 g
4.5 g
✓
45 g
450 g
Correct Answer
4.5 g
At 0.90 g per 100 mL, 500. mL holds 5 times as much: 0.90 g x 5 = 4.5 g of NaCl. 0.45 g treats the percentage as if it were per liter. 45 g is ten times too large, a decimal error. 450 g treats 0.90% as 0.90 g per mL.
Category: Concentration Units
Question 19
The EPA action level for lead in tap water is 15 ppb. What is this concentration expressed in parts per million?
1.5 ppm
15,000 ppm
0.015 ppm
✓
0.15 ppm
Correct Answer
0.015 ppm
One ppm equals 1000 ppb, so 15 ppb divided by 1000 gives 0.015 ppm. 1.5 ppm divides by only 10. 15,000 ppm multiplies by 1000 instead of dividing, though a ppm value must be smaller than the ppb value for the same concentration. 0.15 ppm divides by 100 rather than 1000.
Category: Concentration Units
Question 20
Using Ca = 40.08, N = 14.01, and O = 16.00, what is the molar mass of calcium nitrate, Ca(NO3)2?
102.09 g/mol
116.10 g/mol
150.09 g/mol
164.10 g/mol
✓
Correct Answer
164.10 g/mol
The subscript 2 outside the parentheses doubles everything inside, giving 1 Ca, 2 N, and 6 O: 40.08 + 2(14.01) + 6(16.00) = 164.10 g/mol. 102.09 g/mol ignores the parentheses and counts one nitrate group. 116.10 g/mol doubles the nitrogen but not the oxygen. 150.09 g/mol doubles the oxygen but not the nitrogen.
Category: Formula Mass