### Avogadro Number Calculations II 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 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 unitof K2SO4

Bonus Problem: A sample of HNO3 contains twice as many atoms as there are atoms in 6.840 g of Al2(SO4)3. Calculate the mass of the HNO3 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 1023 formula units/mol) = 1.203882 x 1022 formula units

(1.203882 x 22 formula units) (17 atoms / formula unit) = 2.0466 x 1023 atoms

2) The above calculation done in dimensional analysis style:

 1 mole 6.022 x 1023 form. units 17 atoms 6.840 g x ––––––– x –––––––––––––––––––– x ––––––––– = 2.0466 x 1023 atoms 342.147 g 1 mol 1 form. unit

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

(2.0466 x 1023 atoms) (2) = 4.0932 x 1023 atoms

4) Determine mass of HNO3, done in dimensional analysis style:

 1 form. unit 1 mol 63.0119 g 4.0932 x 1023 atoms x –––––––––– x –––––––––––––––––––– x ––––––– = 8.566 g 5 atoms 6.022 x 1023 form. units 1 mol