AP Chemistry : AP Chemistry

Study concepts, example questions & explanations for AP Chemistry

varsity tutors app store varsity tutors android store

Example Questions

Example Question #701 : Ap Chemistry

How many moles are in 610 atoms of silver?

Possible Answers:

\(\displaystyle 1.0 *10^{-21} mols Ag\)

\(\displaystyle 1.0 1* 10^{-21} mols Ag\)

\(\displaystyle 3.25 * 10^{-11} mols Ag\)

\(\displaystyle 1.01295 * 10^{-21} mols Ag\)

\(\displaystyle 5.1 * 10^{-19} mols Ag\)

Correct answer:

\(\displaystyle 1.0 *10^{-21} mols Ag\)

Explanation:

\(\displaystyle 610 atoms Ag * \frac{1 mol Ag}{6.022 * 10^{23} atoms Ag}= 1.0 * 10^{-21} mols Ag\)

The reason the answer has 2 significant figures, is because the given measurement has 2 significant figures. Remember: in non-decimal numbers, zeros only count if in between non-zero numbers.

Example Question #21 : Atomic Structure And Properties

Which of the following compounds has a formula weight within 74–75 g/mol?

Possible Answers:

C6H12O6

PCl5

BF3

KCl

Correct answer:

KCl

Explanation:

PCl5, BF3, and C6H12O6 are not ionic compounds, and can thus be eliminated because "formula weight" implies that we are dealing with an ionic compound. This leaves only the correct answer, KCl, which can also be verified as correct by calculating the weight. K is 39g, while Cl is 35.5g. Together, this gives a total weight of approximately 74.5g.

Example Question #21 : Atomic Structure And Properties

The empirical formula of an unknown compound is CH2.  Which of the following could not be the molar mass of the unknown compound?

Possible Answers:

56.0 g/mol

84.0 g/mol

42.0 g/mol

70.0 g/mol

16.0 g/mol

Correct answer:

16.0 g/mol

Explanation:

An empirical formula is a way of expressing a chemical formula in the smallest possible ratio. The compound C3H6 has an empirical formula of CH2 because the simplest ratio between this compound's constituent elements is one carbon for every two hydrogens.  This empirical formula can represent a number of compounds, but we know that each compound has a molar mass that is a multiple of the molar mass of the empirical formula. The empirical formula has a molar mass of 14 g/mol so the molar masses of compounds that can be represented by this empirical formula are (28,42,56 . . .).  

Example Question #23 : Atomic Structure And Properties

What is the percent by mass of the anion in lithium iodate, \(\displaystyle LiIO_{4}\)?

Possible Answers:

\(\displaystyle 0.9469\%\)

This compound isn't ionic.

\(\displaystyle 95.36\%\)

\(\displaystyle 96.49\%\)

\(\displaystyle 3.51\%\)

Correct answer:

\(\displaystyle 96.49\%\)

Explanation:

Lithium iodate is comprised of \(\displaystyle Li^{+1}\)and \(\displaystyle (IO_{4})^{-1}\). The anion refers to the negative ion, so we are looking for the percent by mass of \(\displaystyle (IO_{4})^{-1}\).

The total molar mass of lithium iodate is \(\displaystyle 197.84\frac{g}{mol}\).

\(\displaystyle (6.94\frac{g}{mol}Li)+(126.90\frac{g}{mol}I)+4(16.0\frac{g}{mol}O)=197.84\frac{g}{mol}\)

The anion, \(\displaystyle (IO_{4})^{-1}\), contributes 190.9g to the total mass.

\(\displaystyle (126.9\frac{g}{mol}I)+4(16\frac{g}{mol}O)=190.9\frac{g}{mol} IO_{4}^{-1}\)

Find the percent by mass by dividing the mass of the anion by the total molecular mass.

\(\displaystyle \frac{190.9g\ IO_{4}}{197.841g\ total}=96.49\%\)

Example Question #701 : Ap Chemistry

What is the molar mass of \(\displaystyle SOCl_2\)?

Possible Answers:

\(\displaystyle 119\frac{g}{mol}\)

\(\displaystyle 71\frac{g}{mol}\)

\(\displaystyle 110\frac{g}{mol}\)

\(\displaystyle 1\frac{g}{mol}\)

\(\displaystyle 126\frac{g}{mol}\)

Correct answer:

\(\displaystyle 119\frac{g}{mol}\)

Explanation:

To find the molecular weight, you need to add together the atomic masses of each of the atoms. 1 sulfur is 32g/mol, 1 oxygen is 16g/mol, and 2 chlorine is 71g/mol (each Cl atom is 35.5g/mol). These values are taken from the periodic table. Added together, they make 119g/mol.

Example Question #24 : Atomic Structure And Properties

What is the molar mass of \(\displaystyle Ca(NO_3)_2\)?

Possible Answers:

\(\displaystyle 116.1\frac{g}{mol}\)

\(\displaystyle 204.18\frac{g}{mol}\)

\(\displaystyle 164.1\frac{g}{mol}\)

\(\displaystyle 102.09\frac{g}{mol}\)

\(\displaystyle 150.09\frac{g}{mol}\)

Correct answer:

\(\displaystyle 164.1\frac{g}{mol}\)

Explanation:

The chemical formula \(\displaystyle Ca(NO_3)_2\) has a 2 right outside of the parentheses surrounding nitrate, so we need to distribute/multiply that 2 to everything inside the parentheses. That means that we have 1 calcium, two nitrogens, and six oxygens, and we need to add up all those masses to get the molar weight.

\(\displaystyle MW=40.08g+(2\cdot 14.01g)+ (6\cdot 16g)\)

\(\displaystyle MW=164.1\frac{g}{mol}\)

Example Question #26 : Atomic Structure And Properties

What is the molar mass of potassium sulfide?

Possible Answers:

\(\displaystyle 72.15\frac{g}{mol}\)

\(\displaystyle 71.17\frac{g}{mol}\)

\(\displaystyle 142.24\frac{g}{mol}\)

\(\displaystyle 110.27\frac{g}{mol}\)

\(\displaystyle 103.24\frac{g}{mol}\)

Correct answer:

\(\displaystyle 110.27\frac{g}{mol}\)

Explanation:

In order to find the molar mass of potassium sulfide, you must first write out the chemical formula for potassium sulfide. The symbol for the potassium ion is \(\displaystyle K\), and the symbol for sulfur is \(\displaystyle S\). We know that this is an ionic compound because we have a metal involved (potassium) with a nonmetal (sulfur), so we must balance the net charge while making sure both atoms have a full octet. Since potassium is in group I, we know that it can easily lose one electron to fill its outer shell of electrons. Sulfur is in group VI, so we know that it can easily gain two electrons to fill its valence shell of electrons. It would take two potassium atoms to donate one electron each to sulfur to make this compound stable. Thus the chemical formula would be \(\displaystyle K_2S\). Use the periodic table to find the molar mass.

\(\displaystyle MM=(2\cdot 39.10g) +32.07g= 110.27\frac{g}{mol}\)

Example Question #25 : Atomic Structure And Properties

What is the molar mass of ammonium hydroxide?

Possible Answers:

\(\displaystyle 18.042\frac{g}{mol}\)

\(\displaystyle 35.05\frac{g}{mol}\)

\(\displaystyle 32.48\frac{g}{mol}\)

\(\displaystyle 30.01\frac{g}{mol}\)

\(\displaystyle 40.09\frac{g}{mol}\)

Correct answer:

\(\displaystyle 35.05\frac{g}{mol}\)

Explanation:

The formula for ammonium hydroxide is \(\displaystyle NH_4OH\). To find molar mass, add up the individual masses of each element, and remember to multiply that mass by the molar ratio of that element.

\(\displaystyle 14.01+(5\cdot 1.008)+16 = 35.05\frac{g}{mol}\)

Example Question #15 : Moles And Molar Mass

What is the molar mass of oxalic acid?

Possible Answers:

\(\displaystyle 17.05\frac{g}{mol}\)

\(\displaystyle 30.068\frac{g}{mol}\)

\(\displaystyle 27.99\frac{g}{mol}\)

\(\displaystyle 18.058\frac{g}{mol}\)

\(\displaystyle 29.06\frac{g}{mol}\)

Correct answer:

\(\displaystyle 30.068\frac{g}{mol}\)

Explanation:

First, write the formula for oxalic acid: \(\displaystyle H_2C_2O_4\). We know that this formula is correct because when we have an acid with an "-ic" ending and no "hydro-" prefix, that means it's coming from a polyatomic ion with an "-ate" ending. So we know that oxal"ic" acid is derived from the oxal"ate" polyatomic ion. The formula for that ion is \(\displaystyle C_2O_4^{2-}\). Putting a hydrogen atom in front of this ion, and balancing the charges, we get \(\displaystyle H_2C_2O_4\).Then add the atomic masses of each of the individual elements in their respective ratios, to get the molar mass for the whole compound.

\(\displaystyle (2\cdot 1.008)+(2\cdot 12.01)+(2\cdot 16)= 30.068\frac{g}{mol}\)

Example Question #28 : Atomic Structure And Properties

A pure sample of \(\displaystyle KClO_{3}\) is found to contain 71g of chlorine atoms. What is the mass of the sample?

Possible Answers:

\(\displaystyle 250g\)

\(\displaystyle 245 g\)

\(\displaystyle 124g\)

\(\displaystyle 210g\)

\(\displaystyle 180g\)

Correct answer:

\(\displaystyle 245 g\)

Explanation:

First, find how many moles of chlorine are present in the sample:

 \(\displaystyle moles\ Cl = \frac{71g}{35.5\frac{g}{mol}} = 2 mol\) 

Because the ratio of moles of \(\displaystyle Cl\) to moles of \(\displaystyle KClO_{3}\) is 1:1, there are also 2 moles of \(\displaystyle KClO_{3}\). The next step is to find the mass of 2 moles of \(\displaystyle KClO_{3}\).

\(\displaystyle 2mol\ KClO_3 *{122.55\frac{g}{mol}}=245g\)

Learning Tools by Varsity Tutors