## Chemistry: Atoms First (2nd Edition)

a. $$4.01 \times 10^{22} \space atoms \space N$$ b. $$5.97 \times 10^{22} \space atoms \space N$$ c. $$3.67 \times 10^{22} \space atoms \space N$$ d. $$6.54 \times 10^{22} \space atoms \space N$$
a. - Calculate or find the molar mass for $C_2H_5O_2N$: $C_2H_5O_2N$ : ( 1.008 $\times$ 5 )+ ( 12.01 $\times$ 2 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 75.07 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$5.00 \space g \times \frac{1 \space mole}{ 75.07 \space g} = 0.0666 \space mole$$ - Each $C_2H_5O_2N$ has 1 N atoms, thus:$$0.0666 \space mole \space C_2H_5O_2N \times \frac{ 1 \space moles \ N }{1 \space mole \space C_2H_5O_2N } \times \frac{6.022 \times 10^{23} \space atoms \ N }{1 \space mole \space N } = 4.01 \times 10^{22} \space atoms \space N$$ b. - Calculate or find the molar mass for $Mg_3N_2$: $Mg_3N_2$ : ( 24.31 $\times$ 3 )+ ( 14.01 $\times$ 2 )= 100.95 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$5.00 \space g \times \frac{1 \space mole}{ 100.95 \space g} = 0.0495 \underline{3} \space mole$$ - Each $Mg_3N_2$ has 2 N atoms, thus:$$0.0495\underline{3} \space mole \space Mg_3N_2 \times \frac{ 2 \space moles \ N }{1 \space mole \space Mg_3N_2 } \times \frac{6.022 \times 10^{23} \space atoms \ N }{1 \space mole \space N } = 5.97 \times 10^{22} \space atoms \space N$$ c. - Calculate or find the molar mass for $Ca(NO_3)_2$: $Ca(NO_3)_2$ : ( 40.08 $\times$ 1 )+ ( 14.01 $\times$ 2 )+ ( 16.00 $\times$ 6 )= 164.10 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$5.00 \space g \times \frac{1 \space mole}{ 164.10 \space g} = 0.0305 \space mole$$ - Each $Ca(NO_3)_2$ has 2 N atoms, thus:$$0.0305 \space mole \space Ca(NO_3)_2 \times \frac{ 2 \space moles \ N }{1 \space mole \space Ca(NO_3)_2 } \times \frac{6.022 \times 10^{23} \space atoms \ N }{1 \space mole \space N } = 3.67 \times 10^{22} \space atoms \space N$$ d. - Calculate or find the molar mass for $N_2O_4$: $N_2O_4$ : ( 14.01 $\times$ 2 )+ ( 16.00 $\times$ 4 )= 92.02 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$5.00 \space g \times \frac{1 \space mole}{ 92.02 \space g} = 0.0543 \space mole$$ - Each $N_2O_4$ has 2 N atoms, thus:$$0.0543 \space mole \space N_2O_4 \times \frac{ 2 \space moles \ N }{1 \space mole \space N_2O_4 } \times \frac{6.022 \times 10^{23} \space atoms \ N }{1 \space mole \space N } = 6.54 \times 10^{22} \space atoms \space N$$