Question

The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, producing 0.4503 g of SO2, 2.1654 g of CO2, 0.6965 g of H2O, and 0.2109 g of NO. Determine the empirical formula of this compound.

Answer 1

Answer: The empirical formula for the given organic compound is
`C_7H_(11)O_2SN`

Step-by-step explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

`S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO`

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of
`CO_2=0.2829g`

Mass of
`H_2O=0.1159g`

Mass of
`SO_2=0.4503g`

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

• For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide,
`(12)/(44)* 2.1654=0.590g` of carbon will be contained.

• For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water,
`(2)/(18)* 0.6965=0.077g` of hydrogen will be contained.

• For calculating the mass of sulfur:

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide,
`(32)/(64)* 0.4503=0.225g` of sulfur will be contained.

• For calculating the mass of nitrogen:

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide,
`(14)/(30)* 0.2109=0.098g` of nitrogen will be contained.

• Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of Carbon =
`\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=(0.590g)/(12g/mole)=0.049moles`

Moles of Hydrogen =
`\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=(0.077g)/(1g/mole)=0.077moles`

Moles of Oxygen =
`\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=(0.2277g)/(16g/mole)=0.0142moles`

Moles of Sulfur =
`\frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=(0.225g)/(32g/mole)=0.007moles`

Moles of Nitrogen =
`\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=(0.098g)/(14g/mole)=0.007moles`

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon =
`(0.049)/(0.007)=7`

For Hydrogen =
`(0.077)/(0.007)=11`

For Oxygen =
`(0.0142)/(0.007)=2.03\approx 2`

For Sulfur =
`(0.007)/(0.007)=1`

For Nitrogen =
`(0.007)/(0.007)=1`

• Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is
`C_7H_(11)O_2S_1N_1=C_7H_(11)O_2SN`