How Many Atoms or Molecules?

Problems 1 - 10

**Problem #1:** How many atoms of chlorine are in 16.50 g of iron(III) chloride?

**Solution:**

1) Determine moles of FeCl_{3}:

16.50 g / 162.204 g mol¯^{1}= 0.101723755 mol

2) Determine how many formula units of iron(III) chloride are in 0.1017 mol:

0.101723755 mol x 6.022 x 10^{23}= 6.1258 x 10^{22}formula units

3) Determine number of Cl atoms in 6.1258 x 10^{22} formula units of FeCl_{3}:

6.1258 x 10^{22}x 3 = 1.838 x 10^{23}atoms (to 4 sig fig)

4) Set up using dimensional analysis:

1 mol 6.022 x 10 ^{23}3 Cl atoms 16.50 g x –––––––– x –––––––––– x ––––––––– = 1.838 x 10 ^{23}chlorine atoms162.204 g 1 mol 1 FeCl _{3}

**Problem #2:** How much does 100 million atoms of gold weigh?

**Solution:**

1) Determine moles of gold in 1.00 x 10^{8} (we'll assume three sig figs in the 100 million):

1.00 x 10^{8}atoms divided by 6.022 x 10^{23}atoms/mol = 1.660578 x 10¯^{16}mol

2) Determine grams in 1.66 x 10¯^{16} mol of gold:

1.660578 x 10¯^{16}mol times 196.97 g/mol = 3.27 x 10¯^{14}g (to three sig fig)

**Problem #3:** 18.0 g of (NH_{4})_{2}CO_{3} is present. (a) How many oxygen atoms are present? (b) How many hydrogen atoms are present? (c) How many total atoms are present?

**Solution:**

1) Convert mass to moles:

18.0 g / 96.0852 g/mol = 0.18733374 mol

2) Determine how many formula units of ammonium carbonate are present:

(0.18733374 mol) (6.022 x 10^{23}mol¯^{1}) = 1.128124 x 10^{23}formula units

3) Determine oxygen atoms present:

(1.128124 x 10^{23}formula units) (3 O atoms / formula unit) = 3.38 x 10^{23}O atoms

4) Determine hydrogen atoms present:

(1.128124 x 10^{23}formula units) (8 H atoms / formula unit) = 9.02 x 10^{23}H atoms

5) Determine total atoms present:

(1.128124 x 10^{23}formula units) (14 atoms / formula unit) = 1.58 x 10^{24}atomsFor the total, simply count up how many atoms are in the formula unit. In this case, there are 14 total atoms in one formula unit of (NH

_{4})_{2}CO_{3}(two N, eight H, one C and three O).

**Problem #4:** Suppose we knew that there were 8.170 x 10^{20} atoms of O in an unknown sample of KMnO_{4}. How many milligrams would the unknown sample weigh?

**Solution:**

1) Determine how many formula units of KMnO_{4} there are:

8.170 x 10^{20}atoms divided by 4 atoms per formula unit2.0425 x 10

^{20}formula units of KMnO_{4}

2) Determine moles of KMnO_{4}:

2.0425 x 10^{20}formula units divided by 6.022 x 10^{23}formula units/mol = 3.39173 x 10¯^{4}mol

3) Determine grams, then milligrams of KMnO_{4}:

3.39173 x 10¯^{4}mol times 158.032 g/mol = 0.0536 g(0.0536 g) (1000 mg/g) = 53.6 mg

4) Dimensional analysis:

1 formula unit 1 mol 158.032 1000 mg 8.170 x 10 ^{20}atoms x–––––––––––– x –––––––––– x ––––––– x ––––––– = 53.6 mg 4 atoms 6.022 x 10 ^{23}1 mol 1 g

**Problem #5:** A solid sample of cesium sulfate contains 5.780 x 10^{23} cesium ions. How many grams of cesium sulfate must be present?

**Solution:**

1) Determine how many formula units of Cs_{2}SO_{4} must be present:

5.780 x 10^{23}divided by 2 = 2.890 x 10^{23}This is because there are 2 Cs atoms per one cesium sulfate formula unit.

2) Determine how many moles of Cs_{2}SO_{4} are present:

2.890 x 10^{23}divided by 6.022 x 10^{23}mol¯^{1}= 0.479907 mol

3) Determine grams of cesium sulfate:

0.479907 mol times 361.8735 g/mol = 173.7 g

**Problem #6:** A sample of C_{3}H_{8} has 4.64 x 10^{24} H atoms. (a) How many carbon atoms does this sample contain? (b) What is the total mass of the sample?

**Solution:**

1) Convert from hydrogen atoms to C_{3}H_{8} molecules:

4.64 x 10^{24}H atoms divided by 8 H atoms per C_{3}H_{8}molecule = 5.80 x 10^{23}molecules of C_{3}H_{8}

2) C_{3}H_{8} molecules to carbon atoms:

5.80 x 10^{23}molecules times 3 C atoms per molecule = 1.74 x 10^{24}<---answer for (a)

3) Moles of C_{3}H_{8}:

5.80 x 10^{23}molecules divided by 6.022 x 10^{23}molecules / mole = 0.9631352 mol

4) Mass of C_{3}H_{8}:

0.9631352 mol times 44.0962 g/mol = 42.47 gto three sig figs, this is 42.5 g <---answer for (b)

**Problem #7:** If a sample of disulfur hexabromide contains 6.99 x 10^{23} atoms of bromine, what is the mass of the sample?

**Solution:**

1) We need the moles of bromine:

6.99 x 10^{23}atoms divided by 6.022 x 10^{23}atoms/mol = 1.160744 mol or Br atoms

2) Disulfur hexabromide's formula is S_{2}Br_{6}. That means there are 6 moles of Br for every one mole of S_{2}Br_{6}, therefore:

1.160744 mol / 6 = 0.193457 mol of S_{2}Br_{6}present

3) The molar mass of S_{2}Br_{6} is 543.554 g/mol:

543.554 g/mol times 0.193457 mol = 105.15 gto three sig figs, 105 g

4) Dimensional analysis:

1 mol 1 mol 543.554 g 6.99 x 10 ^{23}atoms x–––––––––– x ––––––– x –––––––– = 105 g 6.022 x 10 ^{23}6 mol 1 mol

**Problem #8:** A sample of dinitrogen trioxide contains 0.250 moles of oxygen. How many molecules of the compound are present?

**Solution:**

1) Calculate moles of N_{2}O_{3}:

0.250 mol O times (1 mol N_{2}O_{3}/ 3 mol O) = 0.083333 mol N_{2}O_{3}

2) Calculate molecules of N_{2}O_{3}:

0.083333 mol N_{2}O_{3}times (6.022 x 10^{23}molecules / mol) = 5.02 x 10^{22}molecules (to three sig figs)

**Problem #9:** Determine the number of oxygen atoms in 2.30 g of Al_{2}(SO_{4})_{3}.

**Solution:**

1) Here are the steps:

(a) 2.30 g divided by molar mass of Al_{2}(SO_{4})_{3}= moles of Al_{2}(SO_{4})_{3}(b) moles of Al

_{2}(SO_{4})_{3}times 6.022 x 10^{23}= formula units of Al_{2}(SO_{4})_{3}(c) formula units of Al

_{2}(SO_{4})_{3}times 12 = oxygen atoms in Al_{2}(SO_{4})_{3}

2) Let's build that up in dimensional analysis style:

1 mol 6.022 x 10 ^{23}12 O atoms 2.30 g x ––––––––– x –––––––––– x ––––––– = 4.86 x 10 ^{22}oxygen atoms342.147 g 1 mol 1 ↑ step (a) ↑ ↑ step (b) ↑ ↑ step (c) ↑

**Problem #10:** How many potassium ions are there in 85.0 g of potassium sulfate (K_{2}SO_{4})?

**Solution:**

1 mol 6.022 x 10 ^{23}2 K atoms 85.0 g x ––––––– x –––––––––– x ––––––––––– = 5.87 x 10 ^{23}K atoms174.26 g 1 mol 1 formula unit

of K_{2}SO_{4}

**Problem #11:** A solution of ammonia and water contains 2.10 x 10^{25} water molecules and 8.10 x 10^{24} ammonia molecules. How many total hydrogen atoms are in this solution?

**Solution:**

Ammonia's formula is NH_{3} and water's is H_{2}O. Ammonia has three atoms of H per molecule and water has two atoms of H per molecule.

1) Ammonia's contribution:

8.10 x 10^{24}times 3 = 2.43 x 10^{25}H atoms

2) Water's contribution:

2.10 x 10^{25}times 2 = 4.20 x 10^{25}H atoms

3) Sum them up:

2.43 x 10^{25}+ 4.20 x 10^{25}= 6.63 x 10^{25}H atoms

**Problem #12:** (a) How many water molecules are there in a 4.080 g sample of solid aluminum sulfate octadecahydrate? (b) How many oxygen atoms are there in the 4.080 g sample?

**Solution:**

1) The formula (and molar mass) of aluminum sulfate octadecahydrate is:

Al_{2}(SO_{4})_{3}⋅18H_{2}O666.4134 g/mol

2) Convert 4.080 g of Al_{2}(SO_{4})_{3} **⋅** 18H_{2}O to moles:

4.080 g / 666.4134 g/mol = 0.006122326 mol

3) Determine formula units of Al_{2}(SO_{4})_{3} **⋅** 18H_{2}O in the 4.080 g:

(0.006122326 mol) (6.022 x 10^{23}formula units / mol) = 3.68686 x 10^{21}formula units

4) Every formula unit has 18 water molecules associated with it:

(3.68686 x 10^{21}formula units) (18 water molecules per formula unit) = 6.636 x 10^{22}water molecules (answer to a)

5) In 3.68686 x 10^{21} formula units of Al_{2}(SO_{4})_{3} **⋅** 18H_{2}O, there are a total of 30 oxygen atoms in each formula unit:

(3.68686 x 10^{21}formula units) (30 O atoms per formula unit) = 1.106 x 10^{23}O atoms (answer to b)

**Problem #13:** Determine how many atoms of hydrogen are in 20.0 grams of ammonium chloride.

**Solution #1:**

1) Ammonium chloride = NH_{4}Cl

Mass of one mole of ammonium chloride = 53.4916 gMass of 4 moles of H = 4.0316 g

Fraction H in NH

_{4}Cl = 4.0316 / 53.4916 = 0.075368843

2) To determine the mass of hydrogen in a specific mass of ammonium chloride, multiply by the fraction of H in NH_{4}Cl.

Mass of H = (20.00 g) (0.075368843) = 1.50737686 g

3) Determine moles of hydrogen:

1.50737686 g / 1.008 g/mol = 1.49541355 mol

4) Determine atoms of hydrogen:

(1.49541355 mol) (6.022 x 10^{23}mole¯^{1}) = 9.00 x 10^{23}atoms

**Solution #2:**

1) Convert grams to moles:

20.0 g / 53.4916 g/mol = 0.3738905 mol

2) Convert moles to number of NH_{4}Cl formula units:

(0.3738905 mol) (6.022 x 10^{23}formula units / mole) = 2.2515686 x 10^{23}formula units of NH_{4}Cl

3) There are 4 atoms of hydrogen per formula unit:

(2.2515686 x 10^{23}formula units) (4 atoms / form. unit) = 9.01 x 10^{23}atoms (rounded to three sig figs)

**Problem #14:** How many atoms of mercury are present in 9.70 cubic centimeters of liquid mercury? The density of mercury is 13.55 g/cm^{3}. Answer in units of atoms.

Solution:

1) Determine mass of mercury in 9.70 cm^{3}:

(9.70 cm^{3}) (13.55 g/cm^{3}) = 131.435 g

2) Determine moles of mercury in 131.435 g:

131.435 g / 200.59 g/mol = 0.655242 mol

3) Determine atoms in 0.655242 mol:

(0.655242 mol) (6.022 x 10^{23}atoms/mol) = 3.94 x 10^{23}atoms

**Problem #15:** What is the mass of CH_{4} molecules if they are made from 15.05 x 10^{23} atoms?

**Solution:**

1) In 'x' molecules of methane there are:

'x' atoms of C

'4x' atoms of H

2) From which follows this equation:

x + 4x = 15.05 x 10^{23}x = 3.01 x 10

^{23}atoms of C

3) Since there is 1 atom of C for every 1 molecule of CH_{4}, we have:

3.01 x 10^{23}molecules of CH_{4}

4) Calculate moles of CH_{4}:

3.01 x 10^{23}molecules divided by 6.02 x 10^{23}molecules/mol = 0.500 mole of CH_{4}

5) Calculate mass:

0.500 mol times 16.0426 g/mol = 8.02 g (to three sig figs)

**Bonus Problem:** A sample of HNO_{3} contains twice as many atoms as there are atoms in 6.840 g of Al_{2}(SO_{4})_{3}. Calculate the mass of the HNO_{3} in the sample.

**Solution:**

1) We need to first determine the number of atoms in our sample of aluminum sulfate:

6.840 g / 342.147 g/mol = 0.0199914 mol(0.0199914 mol) (6.022 x 10

^{23}formula units/mol) = 1.203882 x 10^{22}formula units(1.203882 x

^{22}formula units) (17 atoms / formula unit) = 2.0466 x 10^{23}atoms

2) The above calculation done in dimensional analysis style:

1 mole 6.022 x 10 ^{23}form. units17 atoms 6.840 g x ––––––– x –––––––––––––––––––– x ––––––––– = 2.0466 x 10 ^{23}atoms342.147 g 1 mol 1 form. unit

3) The sample of nitric acid will contain twice as many atoms:

(2.0466 x 10^{23}atoms) (2) = 4.0932 x 10^{23}atoms

4) Determine mass of HNO_{3}, done in dimensional analysis style:

1 form. unit 1 mol 63.0119 g 4.0932 x 10 ^{23}atoms x–––––––––– x –––––––––––––––––––– x ––––––– = 8.566 g 5 atoms 6.022 x 10 ^{23}form. units1 mol