Question

one step in making para-aminobenzoic acid,PABA an ingredient in some suntan lotions,involves replacing one of hydrogen atoms in a toluene molecule with an - NO2 group,directly opposite the - CH3 group.calculate the percentage yield when 550g of toluene is added to an excess of nitric acid provides 305g of nitrotouluene priduct.show your work

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Answer 1

The percent yield is 37.2%

Explanation:

Given:

• A reaction between toluene and excess nitric acid yields p-nitro toluene.
• The amount of Toluene = 550 g
• The amount of nitrotoluene produced $m_o$ =  305 g

To find: Percentage Yield

Formula to be used: Percent Yield $= \frac{ m_o}{m_t} \times 100$ where

• $m_o$ is the observed amount of product formed in the reaction
• $m_t$ is the theoretical amount of product formed in the reaction

Solution:

The reaction can be represented as

$Toluene + Nitric acid \rightarrow p-Nitrotoluene$

$C_6H_5CH_3 + HNO_3(excess) \rightarrow C_6H_4(p-NO_2)CH_3$

Step 1: Calculating theoretical yield

For this, we require to calculate the molar masses of each species

• The molar mass of Toluene $= (12\times 6) + (1\times 5) + 12 + 3 =72+5+15 =92$
• The molar mass of p-nitrotoluene$= (12\times 6) + (1\times4) + 12 + 3+14 + (2\times 16) \\=72+4+15+ 14+32 =137$

Thus, 92 g of toluene will produce = 137 g of p-nitro toluene

1 g of toluene will produce = $\frac{137}{92}$ g of p-nitro toluene

So, 550 g toluene will produce = $\frac{137}{92}\times 550$ g  = 819 g of p-nitro toluene

Step 2: Calculating percent yield

• The theoretical amount of product formed in the reaction$, m_t = 819 g$
• The observed amount of product formed in the reaction$, m_o = 305 g$

• Percent Yield $= \frac{305}{819} \times 100$ = 37.2 %