Question

question 23 1 point how many oxygen atoms are in 85.0 milligrams of caffeine? given caffeine is c8h10n4o2 and has a molar mass of 194.0 g/mol. a 2.64e23 atoms b 8.76e - 4 atoms c 6.12e42 atoms d 5.28e20 atoms e none of the above

Explanation:

Step1: Convert mass to grams

$85.0\ mg = 85.0\times10^{- 3}\ g$

Step2: Calculate moles of caffeine

$n=\frac{m}{M}$, where $m = 85.0\times10^{-3}\ g$ and $M = 194.0\ g/mol$. So $n=\frac{85.0\times10^{-3}\ g}{194.0\ g/mol}=4.3814\times10^{-4}\ mol$

Step3: Determine moles of oxygen atoms

In one - mole of $C_8H_{10}N_4O_2$, there are 2 moles of oxygen atoms. So moles of oxygen atoms in $4.3814\times10^{-4}\ mol$ of caffeine is $n_{O}=2\times4.3814\times10^{-4}\ mol = 8.7628\times10^{-4}\ mol$

Step4: Calculate number of oxygen atoms

Using Avogadro's number $N = n\times N_A$, where $N_A=6.022\times10^{23}\ atoms/mol$. So $N = 8.7628\times10^{-4}\ mol\times6.022\times10^{23}\ atoms/mol=5.28\times10^{20}\ atoms$

Answer:

D. 5.28E20 atoms