Question

find the number of atoms of each type present in 3.42 grams of Cane sugar

Answer 1

Solution:

The chemical formula of cane-sugar = $C_{12}H_{22}O_{11}$

We are given the quantity of cane sugar=3.42 grams

We first calculate the molar mass of sugar cane = (12 × 12) + (1 × 22) + (16 × 11)

= (144 + 22 + 176)

= 342 grams

(We know that the atomic mass of C = 12; atomic mass of H = 1; and atomic mass of O = 16)

The number of atoms of C present in cane sugar:

1 mole or 342 gram of cane sugar has 144 g of carbon

3.42 gram of cane sugar $=\frac{(144 \times 3.42)}{342}$

= 1.44 gram of carbon

We also know that 1 mole of anything contains 6.023 \times 10^{23} atoms (Avogadro’s number).

So, 12 g of carbon will have $6.023 \times 10^{23}$

So, 1.44 gram of carbon $=\frac{\left(6.023 \times 10^{23} \times 1.44\right)}{12}$

$=7.226 \times 10^{22} \ atoms$

The number of atoms of H in cane sugar:

1 mole or 342 grams of cane sugar has 22 g of H

3.42 gram of cane sugar $=\frac{(22 \times 3.42)}{342}$

= 0.22 gram of H

We also know that 1 mole of anything contains 6.023 \times 10^{23} atoms (Avogadro’s number).

So 1 g of H will have $6.023 \times 10^{23}$ atoms

So, 0.22 gram will have $=\left(\frac{6.023 \times 10^{23} \times 0.22}{1}\right)$

$=1.325 \times 10^{23}$atoms

The number of atoms of O in cane sugar:

1 mole of 342 gram of cane sugar has 176 g of O

3.42 gram of cane sugar $=\frac{(176 \times 3.42)}{342}$

= 1.76 gram of O

We also know that 1 mole of anything contains 6.023 \times 10^{23} atoms (Avogadro’s number).

So 16 g of O will have 6.023 \times 10^{23} atoms

1.76 grams of O $= \frac{\left(6.023 \times 10^{23} \times 1.76\right)}{16}$

$=6.623 \times 10^{22}$atoms

So, we get

C = $7.226 \times 10^{22}$ atoms

H = $1.325 \times 10^{23}$ atoms

O = $6.623 \times 10^{22}$ atoms

Answer 2

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