3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
Which of the numbered diagrams best represents the arrangement of particles in a small part of a sample of solid sodium chloride, NaCl? Charges are not shown.
Answer and reasoning
ADiagram 1 A student who thinks ionic compounds consist of molecules picks this. The diagram shows separate Na–Cl pairs; solid NaCl has no separate pairs, but an extended array in which each ion is surrounded by ions of the other element.
BDiagram 2 A student who thinks a compound contains molecules of its elements picks this. The chlorine in NaCl is present as separate chloride ions, not as Cl–Cl pairs.
CDiagram 3Correct NaCl is an ionic compound: Na and Cl particles (Na⁺ and Cl⁻ ions) are held in a 1 : 1 ratio in an extended array, each surrounded by particles of the other kind, with no separate molecules. The diagram in which Na and Cl alternate in every row and column shows this.
DDiagram 4 A student who thinks a compound is a mixture of its elements picks this. The diagram shows sodium and chlorine still uncombined, a mixture of the two elements; in NaCl the elements are combined in a fixed ratio.
The table gives the results of analyses of two samples of pure carbon dioxide, CO₂. What is the mass of carbon in sample 2?
Answer and reasoning
A1.20 g A student who thinks every sample of a compound contains the same mass of each element picks this. Every sample has the same proportion of carbon; sample 2 is 2.50 times as massive, so it contains 2.50 times as much carbon.
B3.00 gCorrect Every sample of CO₂ has the same mass ratio of C to CO₂: 1.20/4.40 = 0.273. Sample 2 is 2.50 times as massive as sample 1, so it contains 2.50 × 1.20 g = 3.00 g of carbon.
C3.67 g A student who takes the mass fraction of carbon to be its atom fraction, 1 of the 3 atoms in CO₂, picks this. A C atom has less mass than an O atom, so carbon is only 27.3% of the mass of CO₂.
D7.80 g A student who adds the extra 6.6 g of sample 2 to the mass of carbon picks this. The masses of the elements scale in proportion to the sample mass, by a factor of 2.50, not by adding the same number of grams.
Working By the law of definite proportions, the mass of C is proportional to the mass of CO₂: 1.20 g × (11.0 g/4.40 g) = 1.20 g × 2.50 = 3.00 g. (Check: 12.01/44.01 = 27.3% C; 0.273 × 11.0 g = 3.00 g.)
A student has found the empirical formula of a molecular compound from its percent composition by mass. Which additional quantity does the student need in order to determine the compound's molecular formula?
Answer and reasoning
ANone; the empirical formula is also the molecular formula A student who thinks the empirical formula is the formula of an actual molecule picks this. Different compounds share an empirical formula (CH₂O and C₂H₄O₂), so the empirical formula alone does not give the molecular formula.
BThe total mass of the sample of the compound analyzed A student who thinks percent composition depends on the size of the sample picks this. Every sample of the compound has the same percent composition, so the sample mass gives no information about the size of the molecule.
CThe molar mass of the compound, found by another methodCorrect The molecular formula is a whole-number multiple of the empirical formula. Dividing the compound's molar mass by the mass of the empirical formula gives that multiple; percent composition alone cannot, because CH₂O and C₂H₄O₂ have the same composition.
DThe atomic numbers of the elements in the compound A student who thinks atomic numbers give the masses of atoms picks this. Atomic numbers count protons; the atomic masses were already used to find the empirical formula, and the missing quantity is the molar mass.
Working The molecular formula is n × (empirical formula), where n = (molar mass of compound)/(mass of one empirical formula unit). Only the molar mass is needed in addition to the empirical formula.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
1.3.A.1 Molecule Fix
Molecule
A discrete group of atoms held together by covalent bonds. Molecular substances such as H₂O and CO₂ consist of separate molecules, and their formulas give the number of atoms of each element in one molecule.
Formula unit
The group of ions given by the formula of an ionic compound, such as one Na⁺ and one Cl⁻ for NaCl. An ionic solid is an extended array of ions in this fixed ratio, with no separate molecules.
Pure substance (compound)
A substance with a single kind of particle (one kind of molecule or one formula unit), so its elements are always combined in the same fixed proportions.
Students often think An ionic compound such as NaCl consists of separate molecules, each made of one sodium and one chlorine particle bonded as a pair. In fact No. Solid NaCl is an extended array of Na⁺ and Cl⁻ ions in a 1 : 1 ratio, with each ion surrounded by ions of the opposite charge. NaCl is a formula unit, not a molecule.
Students often think A compound contains molecules of its elements, so solid sodium chloride contains Cl₂ molecules alongside sodium particles. In fact No. In a compound the chlorine is combined with another element: in NaCl it is present as Cl⁻ ions, and no Cl₂ molecules remain. Cl₂ molecules are found in the element chlorine.
1.3.A.2 Law of definite proportions Fix
Law of definite proportions
Every pure sample of a given compound contains its elements in the same ratio by mass, whatever the size or source of the sample. For example, every sample of CO₂ is 27.3% carbon by mass.
Percent composition by mass
The mass of each element in a compound as a percentage of the compound's mass. From the formula: (number of atoms of the element × its atomic mass) ÷ molar mass × 100%.
Students often think The law of definite proportions means that every sample of a compound contains the same mass of each element, whatever the size of the sample. In fact No. Every sample has the same RATIO of element masses (the same percent composition); the actual mass of each element is proportional to the mass of the sample.
Students often think When the mass of a sample of a compound increases, the mass of each element in it increases by the same number of grams as the sample, so differences, not ratios, stay constant. In fact No. The masses of the elements scale in proportion to the sample's mass (they are multiplied by the same factor), not by adding the same amount.
1.3.A.3 Empirical formula Fix
Empirical formula
The formula giving the lowest whole-number ratio of atoms of the elements in a compound, found from the masses (or mass percentages) of the elements by converting each to moles and dividing by the smallest number of moles.
Molecular formula
The formula giving the actual number of atoms of each element in one molecule. It is a whole-number multiple of the empirical formula (C₂H₄O₂ and CH₂O share the empirical formula CH₂O), so finding it from the empirical formula also needs the molar mass.
Students often think The ratio of the masses of the elements in a compound (or their mass percentages) is the same as the ratio of their numbers of atoms in the formula. In fact No. Atoms of different elements have different masses, so the mass ratio equals the atom ratio only if the atoms have equal masses. NaCl has a 1 : 1 atom ratio but contains 1.54 g of chlorine for every 1.00 g of sodium.
Students often think The atomic number of an element can be used as its atomic mass, so masses of elements are converted to moles by dividing by atomic numbers. In fact No. Converting a mass to moles requires the molar mass, which is the atomic mass printed on the periodic table (12.01 for C, 16.00 for O). The atomic number is the number of protons and is not a mass.
7 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 7
Two students buy samples of a white solid labeled 'sodium chloride' from different suppliers. The samples have different masses. The students plan to use elemental analysis to decide whether both samples are the same pure compound. Which question is best aligned with this aim and can be tested by elemental analysis?
Answer and reasoning
AIs the percentage by mass of sodium the same in both samples?Correct By the law of definite proportions, every pure sample of a compound has the same ratio of element masses. Elemental analysis gives the percentage of sodium in each sample, and equal percentages, whatever the sample masses, are what the same compound requires.
BIs the mass of sodium the same in both samples? A student who thinks every sample of a compound contains the same mass of each element picks this. The samples have different masses, so even if both are pure NaCl they contain different masses of sodium.
CAre the masses of sodium and chlorine equal in each sample? A student who thinks the 1 : 1 formula means equal masses of the two elements picks this. A Cl atom has more mass than an Na atom, so pure NaCl contains more chlorine than sodium by mass; the answer would be 'no' for both samples.
DDo both samples contain the elements sodium and chlorine? A student who thinks samples with the same elements must be the same compound picks this. Different compounds, or a compound mixed with another substance, can contain the same elements; only the proportions test whether the samples are the same pure compound.
What mass of oxygen is present in 25.0 g of iron(III) oxide, Fe₂O₃ (molar mass 159.70 g/mol)?
Answer and reasoning
A2.50 g A student who leaves out the subscript of oxygen picks this, using 16.00/159.70. Each formula unit contains three O atoms, so oxygen is 48.00/159.70 = 30.06% of the mass.
B7.51 gCorrect One mole of Fe₂O₃ (159.70 g) contains 3 mol of O atoms (48.00 g), so oxygen is 48.00/159.70 = 30.06% of the mass. 0.3006 × 25.0 g = 7.51 g.
C7.89 g A student who uses atomic numbers (26 for Fe, 8 for O) as masses picks this: 24/76 × 25.0 g = 7.89 g. The atomic masses, 55.85 and 16.00, must be used.
D15.0 g A student who takes the mass fraction of oxygen to be its atom fraction, 3 of the 5 atoms, picks this. An O atom has much less mass than an Fe atom, so oxygen makes up only 30.06% of the mass.
Working Mass fraction of O = (3 × 16.00 g/mol)/(159.70 g/mol) = 48.00/159.70 = 0.3006. Mass of O = 0.3006 × 25.0 g = 7.51 g.
A student heats a sample of copper with excess sulfur in a covered crucible until the reaction is complete and the excess sulfur has burned away. The table shows her data. Which claim about the empirical formula of the product, with its justification, is supported by the data?
Answer and reasoning
ACu₄S, because the mass of Cu is about four times the mass of S A student who takes the mass ratio as the atom ratio picks this. A Cu atom has about twice the mass of an S atom, so a 4 : 1 mass ratio corresponds to a 2 : 1 mole ratio.
BCuS, because copper forms Cu²⁺ ions and sulfur forms S²⁻ ions A student who takes the formula from the most familiar ion charges, ignoring the data, picks this. Copper also forms Cu₂S, and the measured masses give a 2 : 1 mole ratio of Cu to S.
CCu₂S, because dividing each mass by the atomic number gives 2 : 1 A student who uses atomic numbers in place of atomic masses picks this. The ratio 1.27/29 : 0.32/16 happens to be near 2 : 1, but it does not justify the claim; moles must be found with the atomic masses, 63.55 and 32.06.
DCu₂S, because the amount of Cu in moles is twice the amount of SCorrect The product contains 1.27 g of Cu (0.0200 mol) and 0.32 g of S (0.010 mol). The mole ratio Cu : S is 2 : 1, so the empirical formula is Cu₂S.
Working Mass of Cu = 26.89 − 25.62 = 1.27 g; mass of S = 27.21 − 26.89 = 0.32 g. Moles: Cu 1.27/63.55 = 0.0200 mol; S 0.32/32.06 = 0.010 mol. Ratio Cu : S = 2.0 : 1, so the empirical formula is Cu₂S.
A compound with a molar mass of 60.05 g/mol contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is the empirical formula of the compound?
Answer and reasoning
ACH₂OCorrect In 100.0 g there are 3.33 mol C, 6.6 mol H and 3.33 mol O. Dividing by the smallest amount gives C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O.
BC₆HO₈ A student who uses the mass percentages directly as the atom ratio picks this (40.0 : 6.7 : 53.3 ≈ 6 : 1 : 8). The masses must first be divided by the molar masses of the elements.
CCHO A student who divides the percentages by atomic numbers (6, 1 and 8) instead of atomic masses picks this, giving nearly equal numbers. Hydrogen's atomic mass is 1.008 but carbon's is 12.01 and oxygen's 16.00, so there are two H atoms for each C atom.
DC₂H₄O₂ A student who thinks the empirical formula is the formula of an actual molecule picks this, using the molar mass (60.05 g/mol = 2 × 30.03 g/mol) to scale CH₂O up. C₂H₄O₂ is the molecular formula; the empirical formula is the lowest whole-number ratio, CH₂O.
Working In 100.0 g: C 40.0 g ÷ 12.01 g/mol = 3.33 mol; H 6.7 g ÷ 1.008 g/mol = 6.6 mol; O 53.3 g ÷ 16.00 g/mol = 3.33 mol. Dividing by 3.33: C 1.00, H 2.0, O 1.00, so the empirical formula is CH₂O. The molar mass is not needed: it fixes the molecular formula (60.05/30.03 = 2, C₂H₄O₂), not the empirical formula.
The graph shows the mass of element Y in each of four samples of a pure compound of the elements X and Y, plotted against the mass of the sample. Which conclusion is supported by the graph?
Answer and reasoning
ALarger samples contain a greater percentage of Y by mass. A student who confuses the mass of Y with its percentage picks this. Larger samples contain more grams of Y, but the line passes through the origin, so the ratio of Y to sample mass (40%) is the same in every sample.
BOf every 100 atoms in the compound, 40 are Y atoms. A student who takes the mass percentage as the percentage of atoms picks this. The graph gives the mass fraction; the atom ratio depends also on the masses of X and Y atoms, which the graph does not give.
CX exceeds Y by the same number of grams in each sample. A student who reasons additively picks this, expecting the masses of X and Y to rise by equal amounts as the sample gets larger, so that their difference stays fixed. The masses scale in proportion: X is 60% and Y is 40% of each sample, so the difference grows with the sample (0.40 g in the 2.0 g sample, 1.80 g in the 9.0 g sample).
DIn each sample, 40% of the mass is made up of Y.Correct The points lie on a straight line through the origin, so the mass of Y is proportional to the mass of the sample. The slope, 0.40 (for example 3.60 g of Y in 9.0 g), is the mass fraction of Y, the same in every sample, as the law of definite proportions requires.
Working The points lie on a straight line through the origin with slope 4.0 g/10 g = 0.40 (for example 1.80 g/4.5 g = 0.40 and 3.60 g/9.0 g = 0.40). The mass fraction of Y is the same in every sample: 40% Y by mass.
Elements X and Y form two compounds, XY₂ and XY₃. The compound XY₂ is 60.0% X by mass. What is the percent by mass of X in XY₃?
Answer and reasoning
A25.0% A student who takes the percent by mass to be the percentage of atoms picks this, because X is 1 of the 4 atoms in XY₃. The composition of XY₂ shows that an X atom has more mass than a Y atom (1 atom in 3 is X, but X is 60.0% of the mass), so X is more than 25.0% of the mass of XY₃.
B50.0%Correct In XY₂, one X atom accounts for 60.0 parts by mass and two Y atoms for 40.0 parts, so each Y atom accounts for 20.0 parts. One formula unit of XY₃ therefore has 60.0 parts of X in a total of 60.0 + 3(20.0) = 120.0 parts: 60.0/120.0 = 50.0%.
C60.0% A student who thinks two elements combine in the same mass ratio in every compound they form picks this. The constant ratio applies to samples of one compound; XY₃ has more Y for each X than XY₂, so its percentage of X is lower.
D40.0% A student who scales the percentage of Y with its subscript picks this: 40.0% × 3/2 = 60.0% Y, leaving 40.0% X. Adding a Y atom also increases the total mass of the formula unit, so Y is 60.0/120.0 = 50.0% and X is 50.0%.
Working In XY₂ the masses are in the ratio X : 2Y = 60.0 : 40.0, so the mass of one Y atom is 20.0/60.0 = 0.333 of the mass of one X atom. In XY₃: X/(X + 3Y) = 60.0/(60.0 + 3 × 20.0) = 60.0/120.0 = 0.500, or 50.0% X.
The graph shows the mass of iron in samples of a pure oxide of iron, plotted against the mass of the sample. Based on the graph, what is the ratio of Fe atoms to O atoms in the empirical formula of the oxide?
Answer and reasoning
AFe : O = 7 : 3 A student who uses the ratio of the masses as the ratio of the atoms picks this: 7.0 g of Fe to 3.0 g of O. Each mass must be divided by the molar mass of the element; an Fe atom has about 3.5 times the mass of an O atom.
BFe : O = 5 : 7 A student who divides the masses by atomic numbers picks this: 7.0/26 = 0.27 and 3.0/8 = 0.38, a ratio of 1 : 1.4, or 5 : 7. The atomic masses, 55.85 and 16.00, give 0.125 mol and 0.19 mol, a ratio of 2 : 3.
CFe : O = 1 : 2 A student who rounds the mole ratio to whole numbers picks this, writing 1 : 1.5 as 1 : 2. A value of 1.5 is not within experimental error of a whole number; multiplying both values by 2 gives the whole-number ratio 2 : 3.
DFe : O = 2 : 3Correct The line passes through 7.0 g of Fe at 10.0 g of sample, so that sample contains 3.0 g of O. 7.0 g ÷ 55.85 g/mol = 0.125 mol Fe and 3.0 g ÷ 16.00 g/mol = 0.19 mol O, a ratio of 1 : 1.5. Doubling gives the lowest whole-number ratio, 2 : 3.
Working From the graph, a 10.0 g sample contains 7.0 g of Fe, so it contains 10.0 − 7.0 = 3.0 g of O. Fe: 7.0 g ÷ 55.85 g/mol = 0.125 mol. O: 3.0 g ÷ 16.00 g/mol = 0.19 mol. O : Fe = 0.19/0.125 = 1.5, so Fe : O = 1 : 1.5 = 2 : 3 (empirical formula Fe₂O₃).
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
