Question

A 30.5 gram sample of glucose (c6h12o6) contains ________ atoms of oxygen.

Answer 1

First, find the number of moles of glucose in 30.5 g.

You need the molar mass.

Molar mass of C6H12O6 = 6 * 12.01 g/mol + 12 * 1.01 g/mol + 6*16g/mol = 180.2 g/mol

number of moles = mass in grams / molar mass = 30.5 g / 180.2 g/mol = 0.169 mol

Second, use molecular formula

1 mol of C6H12O6 contains 6 moles of O.

Then, 0.169 moles of C6H12O6 contains 6 *0.169 = 1.014 moles of O.

Third, multiply by Avogagro's number:

1.014 moles * 6.022 * 10^23 atoms / mole = 6.106 * 10^ 23 atoms.

Answer: 6.106 * 10 ^23 atoms of oxygen.

Answer 2

There are $6.12 \times 10^{23}$ 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:

1. Convert the mass of glucose into moles of glucose.
2. Find how many moles of oxygen are found in the given moles of glucose.
3. Use Avogadro's number to get the number of atoms present in the given moles of oxygen.

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:

$formula \ mass \ C_6H_{12}O_6 \ = (6)(12.011) \ + \ (12)(1.008) \ + \ (6)(15.999)\\\boxed {formula \ mass \ C_6H_{12}O_6 \ = 180.156}$

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.

$moles \ C_6H_{12}O_6 \ = 30.5 \ g (\frac{1 \ mol}{180.156 \ g})\\\\\boxed { moles \ C_6H_{12}O_6 \ = 0.1693 \ mol}$

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.

$moles \ of \ O \ = 0.1693 \ mol \ C_6H_{12}O_6 \ (\frac{6 \ mol \ O}{1 \ mol \ C_6H_{12}O_6})\\\boxed {moles \ of \ O \ = 1.0158 \ mol}$

STEP 3: Use Avogadro's number to calculate the number of atoms of O

1 mol of O has $6.022 \times 10^{23}$ O atoms

$no. \ of \ O \ atoms \ = 1.0158 \ mol \ O \ \times \frac{6.022 \times 10^{23} \ atoms \ O}{1 \ mol \ O} \\\boxed {no. \ of \ O \ \ atoms \ = 6.117 \times 10^{23} \ atoms}$

Since the given, 30.5 g has 3 significant figures, the final answer must also have 3 significant figures. Therefore,

$\boxed {atoms \ of \ O \ = 6.12 \times 10^{23} \ atoms}}$

LEARN MORE

• Mole Conversion brainly.com/question/12979299
• Stoichiometry brainly.com/question/12979491

Keywords: moles, Avogadro's number