Question

find the number of atoms of each type present in 3.42grams of cane sugar ( C 12 H22O11)

Answer 1

Answer : The number of carbon, hydrogen and oxygen atoms in 3.42 g of $C_{12}H_{22}O_{11}$ is, $7.226\times 10^{22},1.324\times 10^{23}\text{ and }6.624\times 10^{22}$

Solution : Given,

Mass of $C_{12}H_{22}O_{11}$ = 3.42 g

Molar mass of $C_{12}H_{22}O_{11}$ = 342 g/mole

First we have to calculate the moles of $C_{12}H_{22}O_{11}$.

$\text{Moles of }C_{12}H_{22}O_{11}=\frac{\text{Mass of }C_{12}H_{22}O_{11}}{\text{Molar mass of }C_{12}H_{22}O_{11}}=\frac{3.42g}{342g/mole}=0.01moles$

Now we have to calculate the number of atoms of carbon, hydrogen and oxygen.

In the given molecule, there are 12 number of carbon atoms, 22 number of hydrogen atoms and 11 number of oxygen atoms.

As, 1 mole of carbon contains $12\times 6.022\times 10^{23}$ number of hydrogen atoms

So, 0.01 mole of carbon contains $0.01\times 12\times 6.022\times 10^{23}=7.226\times 10^{22}$ number of carbon atoms

and,

As, 1 mole of hydrogen contains $22\times 6.022\times 10^{23}$ number of hydrogen atoms

So, 0.01 mole of hydrogen contains $0.01\times 22\times 6.022\times 10^{23}=1.324\times 10^{23}$ number of hydrogen atoms

and,

As, 1 mole of oxygen contains $11\times 6.022\times 10^{23}$ number of hydrogen atoms

So, 0.01 mole of oxygen contains $0.01\times 11\times 6.022\times 10^{23}=6.624\times 10^{22}$ number of oxygen atoms

Therefore, the number of carbon, hydrogen and oxygen atoms in 3.42 g of $C_{12}H_{22}O_{11}$ is, $7.226\times 10^{22},1.324\times 10^{23}\text{ and }6.624\times 10^{22}$

Answer 2

Answer:

Explanation:

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