Answer
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. $$6.54 \times 10^{22} \space atoms \space N $$
Work Step by Step
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:$$ 0.0543 \space mole \space N_2O_4 \times \frac{ 2 \space moles \ N }{1 \space mole \space N_2O_4 } \times \frac{6.022 \times 10^{23} \space atoms \ N }{1 \space mole \space N } = 6.54 \times 10^{22} \space atoms \space N $$
