3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A student plans to make iron(II) sulfide, FeS, by heating iron with sulfur. She wants to start with equal numbers of Fe atoms and S atoms, but atoms cannot be counted directly in the laboratory. Which procedure is best aligned with her aim?
Answer and reasoning
AWeigh out equal masses of Fe and S, such as 10.0 g of each A student who thinks equal masses contain equal numbers of atoms picks this. An Fe atom has a greater mass than an S atom, so 10.0 g of Fe contains fewer atoms (0.179 mol) than 10.0 g of S (0.312 mol).
BMeasure out equal volumes of powdered Fe and powdered S A student who thinks equal volumes of any two substances contain equal numbers of particles picks this. The atoms of the two solids differ in mass and pack differently, so equal volumes do not give equal numbers of atoms.
CWeigh out masses of Fe and S in the ratio 55.85 g : 32.06 gCorrect One mole of Fe atoms has a mass of 55.85 g and one mole of S atoms has a mass of 32.06 g, and a mole of any element contains 6.022 × 10²³ atoms. Masses in the ratio 55.85 : 32.06 therefore contain equal numbers of atoms, which is how a balance is used to count atoms.
DWeigh out masses of Fe and S in the ratio 26 g : 16 g A student who uses atomic numbers as the relative masses of atoms picks this. 26 and 16 are the numbers of protons in Fe and S atoms; the relative masses of the atoms are the atomic masses, 55.85 and 32.06, so 26 g of Fe (0.466 mol) contains about 7% fewer atoms than 16 g of S (0.499 mol).
The diagram represents two samples of gas, one of O₂ and one of CO₂. Which statement about the two samples shown is correct?
Answer and reasoning
ABoth of the samples of gas have equal total masses. A student who thinks equal numbers of particles have equal masses picks this. A CO₂ molecule (44.01 amu) has a greater mass than an O₂ molecule (32.00 amu), so six CO₂ molecules have the greater mass.
BThe two samples contain equal amounts in moles.Correct Each box contains six molecules. The amount of a substance in moles counts particles (1 mol is 6.022 × 10²³ particles), so equal numbers of molecules are equal amounts in moles, even though the CO₂ molecules are heavier.
CThe CO₂ sample contains the greater amount in moles. A student who treats the amount in moles as a measure of mass picks this. The CO₂ sample has the greater mass, but both boxes contain six molecules, so the amounts in moles are equal.
DThe two samples contain the same number of atoms. A student who equates the number of atoms with the number of molecules picks this. Each O₂ molecule has two atoms and each CO₂ molecule has three, so the boxes contain 12 and 18 atoms.
A student needs to find the mass of 0.250 mol of glucose, C₆H₁₂O₆, whose molar mass is 180.16 g/mol. Which relationship should she use, where m is the mass of the sample, n is the amount in moles, M is the molar mass and NA is Avogadro's number?
Answer and reasoning
Am = n/M A student who swaps the multiplication and division in n = m/M picks this. The units give mol ÷ (g/mol) = mol²/g, not grams; the mass is nM.
Bm = nNA A student who thinks Avogadro's number converts moles to grams picks this. nNA is the number of molecules in the sample, 1.51 × 10²³; the molar mass converts moles to grams.
Cm = M A student who confuses the molar mass with the mass of the sample picks this. 180.16 g is the mass of 1 mol of glucose; 0.250 mol has a quarter of that mass.
Dm = nMCorrect From n = m/M, the mass is m = nM: 0.250 mol × 180.16 g/mol = 45.0 g. The units cancel to grams, which confirms the relationship.
Working n = m/M, so m = nM = (0.250 mol)(180.16 g/mol) = 45.0 g. NA is not needed: it converts moles to a number of molecules, not to a mass.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
1.1.A.1 Mass as a way of counting particles Fix
Mass as a way of counting particles
Particles are far too small and too numerous to count one by one in the laboratory, so chemists count them by weighing: a measured mass, divided by the mass of one mole of the particles, gives the amount of substance and hence the number of particles that react.
Students often think Equal masses of different substances contain equal numbers of atoms or molecules, and equal numbers of particles have equal masses, so mass can be used directly as a count of particles. In fact Not in general. The number of particles in a sample is its mass divided by the average mass of one particle, so equal masses of two substances contain equal numbers of particles only if the substances have the same molar mass. The substance with the lighter particles has more of them.
Students often think Equal volumes of any two substances, including solids, contain equal numbers of particles, so measuring equal volumes gives equal numbers of atoms. In fact Not in general. Particles of different substances differ in size and mass and pack differently, so equal volumes of two solids usually contain different numbers of particles. Chemists find numbers of particles from mass and molar mass.
1.1.A.2 Mole (mol) Fix
Mole (mol)
The unit for the amount of a substance. One mole of a substance contains 6.022 × 10²³ of its constituent particles (atoms, molecules or formula units), so counting in moles lets chemists work with the enormous numbers of particles in laboratory-sized samples.
Avogadro's number (NA)
6.022 × 10²³ mol⁻¹, the number of particles in one mole of a substance. It converts between the amount of a substance in moles, n, and the number of its particles, N: N = n × NA.
Formula unit
The smallest whole-number ratio of ions in an ionic compound, written as its formula (for example, one Ca²⁺ ion and two Cl⁻ ions for CaCl₂). An ionic solid has no separate molecules, so amounts of it are counted in formula units.
Dimensional analysis
Solving a problem by multiplying a quantity by conversion factors (such as 1 mol/44.01 g or 6.022 × 10²³ particles/1 mol) arranged so that the units cancel, leaving the unit of the quantity wanted.
Students often think The amount of a substance in moles is a measure of its mass, so a sample of heavier molecules contains more moles than a sample with the same number of lighter molecules. In fact No. The amount in moles counts particles: equal numbers of molecules are equal amounts in moles, whatever the molecules weigh. The sample of heavier molecules has the greater mass, not the greater amount.
Students often think The number of atoms in a sample equals the number of molecules or formula units, so 1 mol of a compound contains 1 mol of atoms of each element in it. In fact No. Each molecule or formula unit contains the number of atoms shown by its formula, so the number of atoms of an element in a sample is the number of molecules or formula units multiplied by that element's subscript. Only for a monatomic substance such as He are the numbers of atoms and particles equal.
1.1.A.3 Atomic mass unit (amu) Fix
Atomic mass unit (amu)
A unit of mass used for individual atoms and molecules. A ¹²C atom has a mass of exactly 12 amu, and 6.022 × 10²³ amu equals 1 g, which is why the average mass of one particle in amu is numerically equal to the molar mass in grams per mole.
Molar mass (M)
The mass of one mole of a substance, in g/mol. It is numerically equal to the average mass of one of its particles (atom, molecule or formula unit) in amu, and for a compound it is the sum of the atomic masses of the atoms in its formula.
Amount from mass: n = m/M
The amount of a substance in moles, n, equals the mass of the sample, m, divided by the molar mass, M. Combined with NA, it links a measured mass to the number of particles in the sample.
Students often think The atomic number of an element gives the relative mass of its atoms, so it can be used in place of the atomic mass, and an element with a larger atomic number always has heavier atoms. In fact No. The atomic number is the number of protons in the nucleus. The average mass of an atom in amu, which is also the molar mass in g/mol, is the atomic mass printed on the periodic table (55.85 for Fe and 32.06 for S, whose atomic numbers are 26 and 16). Atomic mass usually rises with atomic number but not always: K (atomic number 19) has a smaller atomic mass, 39.10, than Ar (atomic number 18), 39.95.
Students often think One gram of any substance contains Avogadro's number of particles, so the mass of a sample in grams can be used as its amount in moles. In fact No. One mole of a substance has a mass equal to its molar mass in grams, which differs from substance to substance (18.02 g for H₂O, 4.00 g for He). The amount in moles is the mass divided by the molar 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
The table shows the masses of four samples of pure substances. Which sample contains the greatest total number of atoms?
Answer and reasoning
ASample 1Correct 9.00 g ÷ 18.02 g/mol = 0.499 mol of H₂O, and each molecule has three atoms, so the sample contains 1.50 mol of atoms. The other samples contain 0.750 mol (He), 0.666 mol (C₆H₁₂O₆) and 0.626 mol (Ar) of atoms.
BSample 2 A student who compares numbers of molecules instead of atoms picks this. The He sample does contain the most particles (0.750 mol), but each H₂O molecule has three atoms, so the H₂O sample contains 3 × 0.499 = 1.50 mol of atoms.
CSample 3 A student who uses the mass in grams as the amount in moles picks this: 5.00 × 24 atoms per molecule gives the largest product. Dividing by the large molar mass of glucose, 180.16 g/mol, gives only 0.0278 mol of molecules, or 0.666 mol of atoms.
DSample 4 A student who takes the largest mass to contain the most particles picks this. Ar atoms are heavy (39.95 g/mol), so 25.0 g of Ar is only 0.626 mol of atoms, fewer than in the 9.00 g sample of H₂O.
Working Moles of particles, n = m/M, then atoms = n × atoms per particle. He: 3.00 g ÷ 4.00 g/mol = 0.750 mol He, 0.750 mol atoms. H₂O: 9.00 g ÷ 18.02 g/mol = 0.499 mol H₂O × 3 = 1.50 mol atoms. C₆H₁₂O₆: 5.00 g ÷ 180.16 g/mol = 0.0278 mol × 24 = 0.666 mol atoms. Ar: 25.0 g ÷ 39.95 g/mol = 0.626 mol atoms. The H₂O sample has the most atoms, 1.50 mol (9.02 × 10²³ atoms).
The diagram represents a small part of a crystal of an ionic compound formed from the elements M and X. Which statement about a sample containing 1.00 mol of formula units of this compound is correct?
Answer and reasoning
AIt contains 1.00 mol of each of its ions, M and X. A student who thinks 1 mol of an ionic compound contains 1 mol of each ion picks this. The diagram shows two X ions for every M ion, so 1.00 mol of formula units contains 2.00 mol of X ions.
BIt contains 1.00 mol of M ions and 2.00 mol of X ions.Correct The diagram shows twice as many X ions as M ions (12 and 6), so the formula unit is MX₂. One mole of formula units therefore contains 1.00 mol of M ions and 2.00 mol of X ions.
CIt contains 1.00 mol of separate molecules of the compound. A student who thinks ionic compounds consist of molecules picks this. The diagram shows an extended array of ions with no separate groups; the formula unit is the ratio of ions, not a molecule.
DIt contains 1.00 mol of M atoms and 1.00 mol of X₂ molecules. A student who thinks a compound contains molecules of its elements picks this. The crystal contains separate X ions and M ions; no X₂ molecules or neutral M atoms are present.
A student claims that a 10.0 g sample of solid potassium, K, contains more atoms than a 10.0 g sample of argon gas, Ar. Which statement best evaluates the claim?
Answer and reasoning
AThe claim is correct: the average mass of a K atom is less than that of an Ar atom.Correct The average mass of a K atom is 39.10 amu and of an Ar atom 39.95 amu, so the molar masses are 39.10 and 39.95 g/mol. In equal masses, the element with the lighter atoms has more of them: 10.0 g of K is 0.256 mol of atoms and 10.0 g of Ar is 0.250 mol.
BThe claim is incorrect: K has the larger atomic number, so a K atom has the greater mass. A student who takes atomic number as a measure of atomic mass picks this. K has one more proton than Ar, but its average atomic mass, 39.10 amu, is less than that of Ar, 39.95 amu.
CThe claim is incorrect: equal masses of any two elements contain equal numbers of atoms. A student who thinks equal masses contain equal numbers of particles picks this. The number of atoms is the mass divided by the mass of one atom, so equal masses of elements with different atomic masses contain different numbers of atoms.
DThe claim is correct: a gas sample contains fewer atoms than a solid sample of equal mass. A student who thinks the spacing of gas particles means fewer particles picks this. Physical state affects volume, not the number of atoms in a given mass; the claim is correct only because K atoms are lighter than Ar atoms.
How many oxygen atoms are present in 4.10 g of calcium nitrate, Ca(NO₃)₂ (molar mass 164.10 g/mol)?
Answer and reasoning
A1.50 × 10²² A student who takes the number of atoms to equal the number of formula units picks this. 1.50 × 10²² is the number of Ca(NO₃)₂ formula units; each contains six O atoms.
B4.51 × 10²² A student who does not apply the subscript outside the parentheses picks this, counting three O atoms per formula unit. The 2 multiplies the whole NO₃ group, giving six O atoms.
C9.03 × 10²²Correct 4.10 g ÷ 164.10 g/mol = 0.0250 mol of formula units. Each Ca(NO₃)₂ contains two NO₃⁻ ions with three O atoms each, six O atoms in all, so the sample contains 6 × 0.0250 mol × 6.022 × 10²³ mol⁻¹ = 9.03 × 10²² O atoms.
D1.48 × 10²⁵ A student who uses the mass in grams as the amount in moles picks this, multiplying 4.10 by 6 and by NA. The mass must first be divided by the molar mass, 164.10 g/mol.
Working n = 4.10 g ÷ 164.10 g/mol = 0.02498 mol Ca(NO₃)₂. Each formula unit contains 2 × 3 = 6 O atoms: 6 × 0.02498 mol = 0.1499 mol O. Number of O atoms = 0.1499 mol × 6.022 × 10²³ mol⁻¹ = 9.03 × 10²².
Samples of methane, CH₄(g), and oxygen, O₂(g), have equal masses. The total number of atoms in the CH₄ sample is approximately how many times the total number of atoms in the O₂ sample?
Answer and reasoning
A2.0 A student who compares numbers of molecules as if they were numbers of atoms picks this. There are about 2.0 times as many CH₄ molecules, but each has 5 atoms against 2 in O₂, so the ratio of atoms is 5.0.
B2.5 A student who thinks equal masses contain equal numbers of molecules picks this, taking the ratio of atoms as 5/2. A CH₄ molecule has about half the mass of an O₂ molecule, so the CH₄ sample contains about twice as many molecules.
C4.0 A student who uses atomic numbers in place of atomic masses picks this: 'molar masses' of 10 for CH₄ and 16 for O₂ give 16/10 × 5/2 = 4.0. The molar masses are 16.04 and 32.00 g/mol.
D5.0Correct Equal masses contain amounts inversely proportional to molar mass: 32.00/16.04 ≈ 2.0, so there are about twice as many CH₄ molecules as O₂ molecules. Each CH₄ molecule has 5 atoms and each O₂ molecule has 2, so the ratio of atoms is 2.0 × 5/2 = 5.0.
Working For equal masses, n ∝ 1/M: n(CH₄)/n(O₂) = 32.00/16.04 = 1.995 ≈ 2.0. Atoms per molecule: 5 in CH₄, 2 in O₂. Ratio of atoms = 1.995 × 5/2 = 4.99 ≈ 5.0.
What is the average mass, in grams, of one molecule of ethane, C₂H₆?
Answer and reasoning
A1.661 × 10⁻²⁴ g A student who thinks one gram of any substance contains Avogadro's number of particles picks this, dividing 1 g by 6.022 × 10²³. It is 30.07 g of ethane, the molar mass in grams, that contains 6.022 × 10²³ molecules.
B2.989 × 10⁻²³ g A student who uses atomic numbers in place of atomic masses picks this: 2(6) + 6(1) = 18, and 18 ÷ 6.022 × 10²³ = 2.989 × 10⁻²³. The atomic masses, 12.01 and 1.008, give a molar mass of 30.07 g/mol.
C4.993 × 10⁻²³ gCorrect The molar mass of C₂H₆ is 2(12.01) + 6(1.008) = 30.07 g/mol, which is the mass of 6.022 × 10²³ molecules. One molecule therefore has an average mass of 30.07 g ÷ 6.022 × 10²³ = 4.993 × 10⁻²³ g (30.07 amu).
D2.162 × 10⁻²³ g A student who thinks 1 mol of a compound contains 1 mol of atoms of each element picks this, using 12.01 + 1.008 = 13.02 g/mol. One mole of C₂H₆ contains 2 mol of C atoms and 6 mol of H atoms, so its molar mass is 30.07 g/mol.
Working Molar mass of C₂H₆ = 2(12.01) + 6(1.008) = 30.07 g/mol, so one molecule has an average mass of 30.07 amu and 1 mol (6.022 × 10²³ molecules) has a mass of 30.07 g. Mass of one molecule = 30.07 g/mol ÷ 6.022 × 10²³ mol⁻¹ = 4.993 × 10⁻²³ g.
The diagram represents a sample of ammonia gas, NH₃. Each molecule drawn represents 0.0500 mol of NH₃. What is the mass of the sample?
Answer and reasoning
A20.4 g A student who does not distinguish the number of atoms from the number of molecules picks this, counting all 24 atoms drawn as 24 × 0.0500 mol = 1.20 mol of NH₃. The 24 atoms make up six molecules, so the amount of NH₃ is 0.300 mol.
B5.11 gCorrect There are six molecules in the box, so the sample contains 6 × 0.0500 mol = 0.300 mol of NH₃. The molar mass of NH₃ is 14.01 + 3(1.008) = 17.03 g/mol, so the mass is 0.300 mol × 17.03 g/mol = 5.11 g.
C3.00 g A student who uses atomic numbers in place of atomic masses picks this, taking the molar mass as 7 + 3(1) = 10 g/mol. The atomic masses, 14.01 and 1.008, give 17.03 g/mol.
D17.0 g A student who takes the molar mass to be the mass of whatever sample is used picks this. 17.0 g is the mass of 1 mol of NH₃; this sample contains 6 × 0.0500 mol = 0.300 mol.
Working The diagram shows 6 NH₃ molecules, so n = 6 × 0.0500 mol = 0.300 mol. Molar mass of NH₃ = 14.01 + 3(1.008) = 17.03 g/mol. m = nM = (0.300 mol)(17.03 g/mol) = 5.11 g.
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
