## General Chemistry (4th Edition)

(a) To solve the first part, we must use the molar mass conversion factor of Mercury: $$200.59 \: g \:Hg =1 \: mol \:Hg.$$ We are given that there is 3.00 mol Hg so the conversion calculation looks like this: $$mass\: of \: Hg = (3.00\: mol \: Hg) (\frac{200.59 \: g \: Hg}{ 1 \: mol \: Hg}) = 602\: g$$ Note that by including the units in the calculation, we can check that they cancel out properly. (b) For the second part, the molar mass conversion of $Fe(OH)_{3}$ is needed as well as the conversion from moles to particles using Avagadro's constant. These are: $$106.87 \:g \:Fe(OH)_{3} =1 \: mol \:Fe(OH)_{3}$$ $$1 \:mol \:Fe(OH)_{3} = 6.02\times10^{23} \: particles \:Fe(OH)_{3}$$ When these conversions and the given information are combined with the given amount (1.872 $\times\$ 10^{24} particles Fe(OH)_{3}) the units cancel out properly as shown below: $$mass \:of \:Fe(OH)_{3} = (1.872 \times\ 10^{24} \: particles \:Fe(OH)_{3}) (\frac{1 \: mol \:Fe(OH)_{3}}{ 106.87 \:g \:Fe(OH)_{3} })( \frac{1 \:mol \:Fe(OH)_{3} }{6.02\times10^{23} \: particles \:Fe(OH)_{3}}) = 332.2 \:g$$ (c) This part converts from moles to grams using the molar mass conversion for oxygen. Oxygen is naturally a diatomic molecule but in this case we are only concerned with atomic oxygen. Also, this question uses the the isotope of oxygen with a molar mass of ~18.00 in particular. The conversion factor is 18.00 g $^{\:18}O$= 1 mol 18.00$\:g ^{\:18}O$. So, the corresponding calculation is: $$mass\: of \:^{\:18}O \:atoms = (1.00\: mol\: O) (\frac{18.00 \:g ^{\:18}O}{ 1 \:mol \:^{\:18}O}) = 18 \:g$$ d) This part converts from moles to grams using the molar mass conversion for nitrogen. Remember that nitrogen is naturally a diatomic molecule. The conversion factor is 28.01 g$N_{2}$= 1 mol $N_{2}$. So, the corresponding calculation is: $$mass\: of N_{2} = (2.00\: mol N_{2} ) (\frac{28.01 \:g N_{2} }{ 1 \:mol \: N_{2} }) = 56 \:g$$