Problem #6: A sample of C 3H 8 has 4.64 x 10 24 H atoms. This is because there are 2 Cs atoms per one cesium sulfate formula unit.Ģ) Determine how many moles of Cs 2SO 4 are present:Ģ.890 x 10 23 divided by 6.022 x 10 23 mol¯ 1= 0.479907 molĠ.479907 mol times 361.8735 g/mol = 173.7 g How many grams of cesium sulfate must be present?ġ) Determine how many formula units of Cs 2SO 4 must be present:ĥ.780 x 10 23 divided by 2 = 2.890 x 10 23 Problem #5: A solid sample of cesium sulfate contains 5.780 x 10 23 cesium ions. How many milligrams would the unknown sample weigh?ġ) Determine how many formula units of KMnO 4 there are:Ĩ.170 x 10 20 atoms divided by 4 atoms per formula unitĢ.0425 x 10 20 formula units divided by 6.022 x 10 23 formula units/mol = 3.39173 x 10¯ 4 molģ) Determine grams, then milligrams of KMnO 4:ģ.39173 x 10¯ 4 mol times 158.032 g/mol = 0.0536 g Problem #4: Suppose we knew that there were 8.170 x 10 20 atoms of O in an unknown sample of KMnO 4. In this case, there are 14 total atoms in one formula unit of (NH 4) 2CO 3 (two N, eight H, one C and three O). (1.128124 x 10 23 formula units) (14 atoms / formula unit) = 1.58 x 10 24 atomsįor the total, simply count up how many atoms are in the formula unit. (a) How many oxygen atoms are present? (b) How many hydrogen atoms are present? (c) How many total atoms are present?Ģ) Determine how many formula units of ammonium carbonate are present:
Problem #3: 18.0 g of (NH 4) 2CO 3 is present. Problem #2: How much does 100 million atoms of gold weigh?ġ) Determine moles of gold in 1.00 x 10 8 (we'll assume three sig figs in the 100 million):ġ.00 x 10 8 atoms divided by 6.022 x 10 23 atoms/mol = 1.660578 x 10¯ 16 molĢ) Determine grams in 1.66 x 10¯ 16 mol of gold:ġ.660578 x 10¯ 16 mol times 196.97 g/mol = 3.27 x 10¯ 14 g (to three sig fig) We do not charge any extra for these services.ChemTeam: Using Avogadro's Number in Calculations - Problems 1-10
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