There are [tex]6.12 \times 10^{23}[/tex] atoms of oxygen in 30.5 grams of glucose. There is a little over one mole of oxygen in the given mass of glucose, therefore the number of O atoms should be a little over Avogadro's number.
FURTHER EXPLANATION
To get the number of atoms present in 30.5 g of glucose the following steps must be done:
STEP 1: Convert 30.5 g glucose to moles by dividing the given mass by the formula mass (molar mass).
The molar mass of glucose is 180.156 g/mol obtained as follows:
[tex]formula \ mass \ C_6H_{12}O_6 \ = (6)(12.011) \ + \ (12)(1.008) \ + \ (6)(15.999)\\\boxed {formula \ mass \ C_6H_{12}O_6 \ = 180.156}[/tex]
The mass of 1 mol of glucose is equal t its formula mass in g.
Next, set up the equation below to convert the mass of glucose to moles of glucose.
[tex]moles \ C_6H_{12}O_6 \ = 30.5 \ g (\frac{1 \ mol}{180.156 \ g})\\\\\boxed { moles \ C_6H_{12}O_6 \ = 0.1693 \ mol}[/tex]
STEP 2: Determine how many moles of oxygen are in the given moles of glucose. The subscripts of the elements in the chemical formula will give the mole ratio: 1 mole of glucose has 6 moles of oxygen atoms.
[tex]moles \ of \ O \ = 0.1693 \ mol \ C_6H_{12}O_6 \ (\frac{6 \ mol \ O}{1 \ mol \ C_6H_{12}O_6})\\\boxed {moles \ of \ O \ = 1.0158 \ mol}[/tex]
STEP 3: Use Avogadro's number to calculate the number of atoms of O
1 mol of O has [tex]6.022 \times 10^{23}[/tex] O atoms
[tex]no. \ of \ O \ atoms \ = 1.0158 \ mol \ O \ \times \frac{6.022 \times 10^{23} \ atoms \ O}{1 \ mol \ O} \\\boxed {no. \ of \ O \ \ atoms \ = 6.117 \times 10^{23} \ atoms}[/tex]
Since the given, 30.5 g has 3 significant figures, the final answer must also have 3 significant figures. Therefore,
[tex]\boxed {atoms \ of \ O \ = 6.12 \times 10^{23} \ atoms}}[/tex]
LEARN MORE
Keywords: moles, Avogadro's number
