1. Determine the number of atoms in 1.75 mol Al
2. Calculate the number of formula units in 25.0 mol NaNO3
3. Given 6.25 mol H O2 determine the number of molecules.
DIRECTIONS: A. How many atoms are in each of the following samples?
a atoms Ne
b molecules CH4
1. What is the percent composition of the following compounds?
a Naphtalene (C10 H8)
b Sucrose (C12 H22 O11)
c Alu um sulfate Al2(SO4)3
1. N(Al) = n(Al) × NA = 1.75 mol × 6.022 × 1023 = 1.05 × 1024 atoms
2. N(NaNO3) = n(NaNO3) × NA = 25.0 mol × 6.022 × 1023 = 1.50 × 1025 formula units
3. N(H2O) = n(H2O) × NA = 6.25 mol × 6.022 × 1023 = 3.76 × 1024 molecules
4. a. One mole of Ne contains 6.022 × 1023 atoms;
b. one mole of CH4 contains 6.022 × 1023 molecules and 3.01 × 1024 atoms
5. a. Mr(C10H8) = 128 g/mol; %(C) = (12 g/mol × 10) / 128 g/mol = 0.9375 = 93.75 %; %(H) = 100 % - 93.75 % = 6.25 %
b. Mr(C12H22O11) = 342 g/mol; %(C) = (12 g/mol × 12) / 342 g/mol = 0.421 = 42.1 %; %(H) = 22 g/mol / 342 g/mol = 0.064 = 6.4 %; %(O) = 100 % - 42.1 % - 6.4 % = 51.5 %
c. Mr(Al2(SO4)3) = 342 g/mol; %(Al) = (27 g/mol × 2) / 342 g/mol = 0.158 = 15.8 %; %(S) = (32 g/mol × 3) / 342 g/mol = 0.281 = 28.1 %; %(O) = 100 % - 15.8 % - 28.1 % = 56.1 %.
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