Question

20) A compound was analyzed and found to be 47% Carbon, 11% Hydrogen and 42% Oxygen, What is its Empirical Formula? What is its molecular formula if its molecular weight is 304 g/mole?

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question: in.....typed format...

$\Rightarrow$Answer:

1. Empirical formula of compound:  C₄O₄H (i.e. a rough estimate, for the given compound...)
2. Molecular formula of the compound will be:   C₁₂H₃₃O₈...

$\Rightarrow$Explanation:

Following is the complete Explanation: for the above answer: in....typed format...

• Given:

The given compound is found to be having the following: ( assumed to be mass percentages)

1. Percentage of carbon ( C ) = 47%  
2. Percentage of hydrogen ( H ) = 11%
3. Percentage of oxygen ( O ) = 42%   
4. Molecular weight of the compound = 304.0 g/mol ( grams per mole )

• ​​​​​​​Step - 1:

​​​​​​​Using the above percentages, we will get the following rough ratios of the atoms in the compound:

1. Ratio of hydrogen ( H ) to carbon (C): = 11% /  47% = 1: 4.27  $\approx$  1 : 4
2. Ratio of hydrogen ( H ) to oxygen (O): = 11% : 42% = 1: 3.818 $\approx$ 1: 4

Therefore, using the given atomic ratios, the rough estimate of the Empirical formula of the compound will be the following:

$\Rightarrow$ Empirical formula:   C₄O₄H (i.e. a rough estimate, for the given compound...)

• Step - 2:

Now, considering the given Ratios: to be mass ratios of the elements: in the compound: we will be able to calculate the following:

1. Mass of carbon ( C ) in the compound = ( molar mass) x ( mass percentage)

= (304 g/mol ) x ( 47%)

= 142.88 g (grams)...per mole of compound.

1. Mass of hydrogen ( H ) = ( 304.0 g/mol ) x ( 11%) = 33.44 g (grams)...per mole of compound.
2. Mass of oxygen ( O ) = ( 304.0 g/mol ) x ( 42% ) = 127.68 g (grams)...per mole of compound.

• ​​​​​​​Step - 3:

​​​​​​​We know, the following atomic masses: in g/mol , of the given elements:

1. Atomic mass of carbon ( C ) = 12.011 g/mol
2. Atomic mass of hydrogen ( H ) = 1.008 g/mol
3. Atomic mass of oxygen ( O ) = 16.0 g/mol

• ​​​​​​​Step - 4:

​​​​​​​Using the above masses, and the above atomic masses: we can calculate the following:

1. Number of atoms of carbon ( C ) = (142.88 g ) / ( 12.011 g/atom ) = 11.89 $\approx$   12.0
2. Number of atoms of hydrogen ( H ) = (33.44 g ) / ( 1.008 g/atom ) = 33.17  $\approx$  33.0
3. Number of atoms of oxygen ( O ) = (127.68 g ) / ( 16.00 g/atom ) = 7.98  $\approx$  8.0

​​​​​​​Therefore, the molecular formula of the compound will be the following:

$\Rightarrow$ molecular formula: C₁₂H₃₃O₈...

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