Avogadro Number Calculations II
How Many Atoms or Molecules?
Problems 1 - 10

Ten Examples

Mole Table of Contents


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

Solution:

1) Determine moles of FeCl3:

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 1023 = 6.1258 x 1022 formula units

3) Determine number of Cl atoms in 6.1258 x 1022 formula units of FeCl3:

6.1258 x 1022 x 3 = 1.838 x 1023 atoms (to 4 sig fig)

4) Set up using dimensional analysis:

  1 mol   6.022 x 1023   3 Cl atoms  
16.50 g x –––––––– x –––––––––– x ––––––––– = 1.838 x 1023 chlorine atoms
  162.204 g   1 mol   1 FeCl3  

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

Solution:

1) Determine moles of gold in 1.00 x 108 (we'll assume three sig figs in the 100 million):

1.00 x 108 atoms divided by 6.022 x 1023 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 (NH4)2CO3 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 1023 mol¯1) = 1.128124 x 1023 formula units

3) Determine oxygen atoms present:

(1.128124 x 1023 formula units) (3 O atoms / formula unit) = 3.38 x 1023 O atoms

4) Determine hydrogen atoms present:

(1.128124 x 1023 formula units) (8 H atoms / formula unit) = 9.02 x 1023 H atoms

5) Determine total atoms present:

(1.128124 x 1023 formula units) (14 atoms / formula unit) = 1.58 x 1024 atoms

For 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 (NH4)2CO3 (two N, eight H, one C and three O).


Problem #4: Suppose we knew that there were 8.170 x 1020 atoms of O in an unknown sample of KMnO4. How many milligrams would the unknown sample weigh?

Solution:

1) Determine how many formula units of KMnO4 there are:

8.170 x 1020 atoms divided by 4 atoms per formula unit

2.0425 x 1020 formula units of KMnO4

2) Determine moles of KMnO4:

2.0425 x 1020 formula units divided by 6.022 x 1023 formula units/mol = 3.39173 x 10¯4 mol

3) Determine grams, then milligrams of KMnO4:

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 1020 atoms x –––––––––––– x –––––––––– x ––––––– x ––––––– = 53.6 mg
  4 atoms   6.022 x 1023   1 mol   1 g  

Problem #5: A solid sample of cesium sulfate contains 5.780 x 1023 cesium ions. How many grams of cesium sulfate must be present?

Solution:

1) Determine how many formula units of Cs2SO4 must be present:

5.780 x 1023 divided by 2 = 2.890 x 1023

This is because there are 2 Cs atoms per one cesium sulfate formula unit.

2) Determine how many moles of Cs2SO4 are present:

2.890 x 1023 divided by 6.022 x 1023 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 C3H8 has 4.64 x 1024 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 C3H8 molecules:

4.64 x 1024 H atoms divided by 8 H atoms per C3H8 molecule = 5.80 x 1023 molecules of C3H8

2) C3H8 molecules to carbon atoms:

5.80 x 1023 molecules times 3 C atoms per molecule = 1.74 x 1024 <---answer for (a)

3) Moles of C3H8:

5.80 x 1023 molecules divided by 6.022 x 1023 molecules / mole = 0.9631352 mol

4) Mass of C3H8:

0.9631352 mol times 44.0962 g/mol = 42.47 g

to three sig figs, this is 42.5 g <---answer for (b)


Problem #7: If a sample of disulfur hexabromide contains 6.99 x 1023 atoms of bromine, what is the mass of the sample?

Solution:

1) We need the moles of bromine:

6.99 x 1023 atoms divided by 6.022 x 1023 atoms/mol = 1.160744 mol or Br atoms

2) Disulfur hexabromide's formula is S2Br6. That means there are 6 moles of Br for every one mole of S2Br6, therefore:

1.160744 mol / 6 = 0.193457 mol of S2Br6 present

3) The molar mass of S2Br6 is 543.554 g/mol:

543.554 g/mol times 0.193457 mol = 105.15 g

to three sig figs, 105 g

4) Dimensional analysis:

  1 mol   1 mol   543.554 g  
6.99 x 1023 atoms x –––––––––– x ––––––– x –––––––– = 105 g
  6.022 x 1023   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 N2O3:

0.250 mol O times (1 mol N2O3 / 3 mol O) = 0.083333 mol N2O3

2) Calculate molecules of N2O3:

0.083333 mol N2O3 times (6.022 x 1023 molecules / mol) = 5.02 x 1022 molecules (to three sig figs)

Problem #9: Determine the number of oxygen atoms in 2.30 g of Al2(SO4)3.

Solution:

1) Here are the steps:

(a) 2.30 g divided by molar mass of Al2(SO4)3 = moles of Al2(SO4)3

(b) moles of Al2(SO4)3 times 6.022 x 1023 = formula units of Al2(SO4)3

(c) formula units of Al2(SO4)3 times 12 = oxygen atoms in Al2(SO4)3

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

  1 mol   6.022 x 1023   12 O atoms  
2.30 g x ––––––––– x –––––––––– x ––––––– = 4.86 x 1022 oxygen atoms
  342.147 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 (K2SO4)?

Solution:

  1 mol   6.022 x 1023   2 K atoms  
85.0 g x ––––––– x –––––––––– x ––––––––––– = 5.87 x 1023 K atoms
  174.26 g   1 mol   1 formula unit
of K2SO4
 

Problem #11: A solution of ammonia and water contains 2.10 x 1025 water molecules and 8.10 x 1024 ammonia molecules. How many total hydrogen atoms are in this solution?

Solution:

Ammonia's formula is NH3 and water's is H2O. Ammonia has three atoms of H per molecule and water has two atoms of H per molecule.

1) Ammonia's contribution:

8.10 x 1024 times 3 = 2.43 x 1025 H atoms

2) Water's contribution:

2.10 x 1025 times 2 = 4.20 x 1025 H atoms

3) Sum them up:

2.43 x 1025 + 4.20 x 1025 = 6.63 x 1025 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:

Al2(SO4)3 18H2O

666.4134 g/mol

2) Convert 4.080 g of Al2(SO4)3 18H2O to moles:

4.080 g / 666.4134 g/mol = 0.006122326 mol

3) Determine formula units of Al2(SO4)3 18H2O in the 4.080 g:

(0.006122326 mol) (6.022 x 1023 formula units / mol) = 3.68686 x 1021 formula units

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

(3.68686 x 1021 formula units) (18 water molecules per formula unit) = 6.636 x 1022 water molecules (answer to a)

5) In 3.68686 x 1021 formula units of Al2(SO4)3 18H2O, there are a total of 30 oxygen atoms in each formula unit:

(3.68686 x 1021 formula units) (30 O atoms per formula unit) = 1.106 x 1023 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 = NH4Cl

Mass of one mole of ammonium chloride = 53.4916 g

Mass of 4 moles of H = 4.0316 g

Fraction H in NH4Cl = 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 NH4Cl.

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 1023 mole¯1) = 9.00 x 1023 atoms

Solution #2:

1) Convert grams to moles:

20.0 g / 53.4916 g/mol = 0.3738905 mol

2) Convert moles to number of NH4Cl formula units:

(0.3738905 mol) (6.022 x 1023 formula units / mole) = 2.2515686 x 1023 formula units of NH4Cl

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

(2.2515686 x 1023 formula units) (4 atoms / form. unit) = 9.01 x 1023 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/cm3. Answer in units of atoms.

Solution:

1) Determine mass of mercury in 9.70 cm3:

(9.70 cm3) (13.55 g/cm3) = 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 1023 atoms/mol) = 3.94 x 1023 atoms

Bonus Problem: What is the mass of CH4 molecules if they are made from 15.05 x 1023 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 1023

x = 3.01 x 1023 atoms of C

3) Since there is 1 atom of C for every 1 molecule of CH4, we have:

3.01 x 1023 molecules of CH4

4) Calculate moles of CH4:

3.01 x 1023 molecules divided by 6.02 x 1023 molecules/mol = 0.500 mole of CH4

5) Calculate mass:

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

Ten Examples

Mole Table of Contents