Question

2. A sample of table sugar (sucrose, C12H22O11) has a mass of 1.202 g.

Answer 1

Explanation:

Given mass of sugar = 1.202 g

Molar mass of the sugar , $C_{12}H_{22}O_{11}$ = 342.29 g/mol

a) Number of moles of sugar

$\text{number of moles}=\frac{\text{given mass of sugar}}{\text{molar mass of sugar}}=\frac{1.202 g}{342.29g/mol}=0.0035 mol$

Number of moles of sugar in 1.202 grams is 0.0035 moles

b)Moles of each element in $C_{12}H_{22}O_{11}$

In a molecular formula of sugar there are 12 carbon, 22 hydrogen and 11 oxygen atoms

Moles of an element in molecule =

number of atoms of that element in a molecular formula × number of moles of compound

Moles of carbon = 12 × 0.0035 = 0.042 moles

Similarly

Moles of hydrogen = 22  × 0.0035 = 0.077 moles

Moles of oxygen =11 × 0.0035 = 0.0385 moles

c) Number of atoms of each type in $C_{12}H_{22}O_{11}$

$\text{Number of atoms}=\text{number of moles of that atom in a molecular formula}\times \text{Avogadro number}$

Number of carbon atoms =$0.042 mole\times 6.022 \times 10^{23}=2.52\times 10^{22} atoms$

Number of hydrogen =$0.077mole\times 6.022 \times 10^{23}=4.63\times 10^{22} atoms$

Number of oxygen atoms = $0.0385 mole\times 6.022 \times 10^{23}=2.31\times 10^{22} atoms$

Answer 2

a. Calculate the number of moles of C12H22O11 contained in the sample and record
 1.202 g C12H22O11 ( 1 mol / 342.30 g ) = 0.0035 mol C12H22O11

b. Calculate the moles of each element in C12H22O11

0.0035 C12H22O11 ( 12 mol C / 1 mol C12H22O11 ) = 0.042 mol C
0.0035 C12H22O11 ( 22 mol H / 1 mol C12H22O11 ) = 0.077 mol H
0.0035 C12H22O11 ( 11 mol C / 1 mol C12H22O11 ) = 0.0385 mol O

c. Calculate the number of atoms of each type in C12H22O11

0.042 mol C ( 6.022x10^23 atoms / 1 mol ) = 2.53x10^22 atoms C
0.077 mol H ( 6.022x10^23 atoms  / 1 mol  ) = 4.64x10^22 atoms H
0.0385 mol O ( 6.022x10^23 atoms  / 1 mol  ) = 2.32x10^22 atoms O