What Mass (in Grams) Of Aspirin $\left( C_9H_8O_4 \right$\] Is Produced From 77.7 G Of $C_7H_6O_3$, Assuming A $95.0\%$ Yield From The Reaction Below?$\[ C_7H_6O_3(s) + C_4H_6O_3(l) \rightarrow C_9H_8O_4(s) +

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What mass of aspirin is produced from 77.7 g of C7H6O3?

Understanding the Chemical Reaction

The given chemical reaction is:

C7H6O3(s)+C4H6O3(l)→C9H8O4(s)+other products{ C_7H_6O_3(s) + C_4H_6O_3(l) \rightarrow C_9H_8O_4(s) + \text{other products} }

However, the reaction is incomplete, and we are only interested in the production of aspirin (C9H8O4C_9H_8O_4). To determine the mass of aspirin produced, we need to know the stoichiometry of the reaction.

Balancing the Chemical Equation

To balance the chemical equation, we need to know the number of moles of each reactant and product. Let's assume the balanced equation is:

C7H6O3(s)+C4H6O3(l)→C9H8O4(s)+other products{ C_7H_6O_3(s) + C_4H_6O_3(l) \rightarrow C_9H_8O_4(s) + \text{other products} }

However, we don't have enough information to balance the equation. We can assume that the reaction is a simple combination reaction, where one mole of C7H6O3C_7H_6O_3 reacts with one mole of C4H6O3C_4H_6O_3 to produce one mole of C9H8O4C_9H_8O_4.

Calculating the Number of Moles of C7H6O3

To calculate the number of moles of C7H6O3C_7H_6O_3, we need to know its molar mass. The molar mass of C7H6O3C_7H_6O_3 is:

MC7H6O3=7(12.01 g/mol)+6(1.008 g/mol)+3(16.00 g/mol){ M_{C_7H_6O_3} = 7(12.01\, \text{g/mol}) + 6(1.008\, \text{g/mol}) + 3(16.00\, \text{g/mol}) }

MC7H6O3=84.07 g/mol+6.048 g/mol+48.00 g/mol{ M_{C_7H_6O_3} = 84.07\, \text{g/mol} + 6.048\, \text{g/mol} + 48.00\, \text{g/mol} }

MC7H6O3=138.12 g/mol{ M_{C_7H_6O_3} = 138.12\, \text{g/mol} }

Now, we can calculate the number of moles of C7H6O3C_7H_6O_3:

nC7H6O3=mC7H6O3MC7H6O3{ n_{C_7H_6O_3} = \frac{m_{C_7H_6O_3}}{M_{C_7H_6O_3}} }

nC7H6O3=77.7 g138.12 g/mol{ n_{C_7H_6O_3} = \frac{77.7\, \text{g}}{138.12\, \text{g/mol}} }

nC7H6O3=0.563 mol{ n_{C_7H_6O_3} = 0.563\, \text{mol} }

Calculating the Number of Moles of C9H8O4

Since the reaction is assumed to be a simple combination reaction, the number of moles of C9H8O4C_9H_8O_4 produced is equal to the number of moles of C7H6O3C_7H_6O_3 reacted:

nC9H8O4=nC7H6O3{ n_{C_9H_8O_4} = n_{C_7H_6O_3} }

nC9H8O4=0.563 mol{ n_{C_9H_8O_4} = 0.563\, \text{mol} }

Calculating the Mass of C9H8O4 Produced

To calculate the mass of C9H8O4C_9H_8O_4 produced, we need to know its molar mass. The molar mass of C9H8O4C_9H_8O_4 is:

MC9H8O4=9(12.01 g/mol)+8(1.008 g/mol)+4(16.00 g/mol){ M_{C_9H_8O_4} = 9(12.01\, \text{g/mol}) + 8(1.008\, \text{g/mol}) + 4(16.00\, \text{g/mol}) }

MC9H8O4=108.09 g/mol+8.064 g/mol+64.00 g/mol{ M_{C_9H_8O_4} = 108.09\, \text{g/mol} + 8.064\, \text{g/mol} + 64.00\, \text{g/mol} }

MC9H8O4=180.15 g/mol{ M_{C_9H_8O_4} = 180.15\, \text{g/mol} }

Now, we can calculate the mass of C9H8O4C_9H_8O_4 produced:

mC9H8O4=nC9H8O4×MC9H8O4{ m_{C_9H_8O_4} = n_{C_9H_8O_4} \times M_{C_9H_8O_4} }

mC9H8O4=0.563 mol×180.15 g/mol{ m_{C_9H_8O_4} = 0.563\, \text{mol} \times 180.15\, \text{g/mol} }

mC9H8O4=101.5 g{ m_{C_9H_8O_4} = 101.5\, \text{g} }

Calculating the Mass of Aspirin Produced with 95.0% Yield

Since the yield of the reaction is 95.0%, the mass of aspirin produced is:

maspirin=mC9H8O41.00×0.950{ m_{\text{aspirin}} = \frac{m_{C_9H_8O_4}}{1.00} \times 0.950 }

maspirin=101.5 g1.00×0.950{ m_{\text{aspirin}} = \frac{101.5\, \text{g}}{1.00} \times 0.950 }

maspirin=96.7 g{ m_{\text{aspirin}} = 96.7\, \text{g} }

Therefore, the mass of aspirin produced from 77.7 g of C7H6O3C_7H_6O_3 is approximately 96.7 g.

Conclusion

In this problem, we calculated the mass of aspirin produced from 77.7 g of C7H6O3C_7H_6O_3 using the given chemical reaction and a 95.0% yield. We found that the mass of aspirin produced is approximately 96.7 g. This calculation demonstrates the importance of understanding chemical reactions and stoichiometry in chemistry.

References

Glossary

  • Molar mass: The mass of one mole of a substance.
  • Mole: A unit of measurement for the amount of a substance.
  • Stoichiometry: The study of the quantitative relationships between reactants and products in chemical reactions.
  • Yield: The percentage of the desired product obtained from a chemical reaction.
    Q&A: What mass of aspirin is produced from 77.7 g of C7H6O3?

Frequently Asked Questions

Q: What is the chemical reaction for producing aspirin?

A: The chemical reaction for producing aspirin is:

C7H6O3(s)+C4H6O3(l)→C9H8O4(s)+other products{ C_7H_6O_3(s) + C_4H_6O_3(l) \rightarrow C_9H_8O_4(s) + \text{other products} }

However, the reaction is incomplete, and we are only interested in the production of aspirin (C9H8O4C_9H_8O_4).

Q: How do I calculate the number of moles of C7H6O3?

A: To calculate the number of moles of C7H6O3C_7H_6O_3, you need to know its molar mass. The molar mass of C7H6O3C_7H_6O_3 is:

MC7H6O3=7(12.01 g/mol)+6(1.008 g/mol)+3(16.00 g/mol){ M_{C_7H_6O_3} = 7(12.01\, \text{g/mol}) + 6(1.008\, \text{g/mol}) + 3(16.00\, \text{g/mol}) }

MC7H6O3=84.07 g/mol+6.048 g/mol+48.00 g/mol{ M_{C_7H_6O_3} = 84.07\, \text{g/mol} + 6.048\, \text{g/mol} + 48.00\, \text{g/mol} }

MC7H6O3=138.12 g/mol{ M_{C_7H_6O_3} = 138.12\, \text{g/mol} }

Now, you can calculate the number of moles of C7H6O3C_7H_6O_3:

nC7H6O3=mC7H6O3MC7H6O3{ n_{C_7H_6O_3} = \frac{m_{C_7H_6O_3}}{M_{C_7H_6O_3}} }

nC7H6O3=77.7 g138.12 g/mol{ n_{C_7H_6O_3} = \frac{77.7\, \text{g}}{138.12\, \text{g/mol}} }

nC7H6O3=0.563 mol{ n_{C_7H_6O_3} = 0.563\, \text{mol} }

Q: How do I calculate the number of moles of C9H8O4?

A: Since the reaction is assumed to be a simple combination reaction, the number of moles of C9H8O4C_9H_8O_4 produced is equal to the number of moles of C7H6O3C_7H_6O_3 reacted:

nC9H8O4=nC7H6O3{ n_{C_9H_8O_4} = n_{C_7H_6O_3} }

nC9H8O4=0.563 mol{ n_{C_9H_8O_4} = 0.563\, \text{mol} }

Q: How do I calculate the mass of C9H8O4 produced?

A: To calculate the mass of C9H8O4C_9H_8O_4 produced, you need to know its molar mass. The molar mass of C9H8O4C_9H_8O_4 is:

MC9H8O4=9(12.01 g/mol)+8(1.008 g/mol)+4(16.00 g/mol){ M_{C_9H_8O_4} = 9(12.01\, \text{g/mol}) + 8(1.008\, \text{g/mol}) + 4(16.00\, \text{g/mol}) }

MC9H8O4=108.09 g/mol+8.064 g/mol+64.00 g/mol{ M_{C_9H_8O_4} = 108.09\, \text{g/mol} + 8.064\, \text{g/mol} + 64.00\, \text{g/mol} }

MC9H8O4=180.15 g/mol{ M_{C_9H_8O_4} = 180.15\, \text{g/mol} }

Now, you can calculate the mass of C9H8O4C_9H_8O_4 produced:

mC9H8O4=nC9H8O4×MC9H8O4{ m_{C_9H_8O_4} = n_{C_9H_8O_4} \times M_{C_9H_8O_4} }

mC9H8O4=0.563 mol×180.15 g/mol{ m_{C_9H_8O_4} = 0.563\, \text{mol} \times 180.15\, \text{g/mol} }

mC9H8O4=101.5 g{ m_{C_9H_8O_4} = 101.5\, \text{g} }

Q: How do I calculate the mass of aspirin produced with 95.0% yield?

A: Since the yield of the reaction is 95.0%, the mass of aspirin produced is:

maspirin=mC9H8O41.00×0.950{ m_{\text{aspirin}} = \frac{m_{C_9H_8O_4}}{1.00} \times 0.950 }

maspirin=101.5 g1.00×0.950{ m_{\text{aspirin}} = \frac{101.5\, \text{g}}{1.00} \times 0.950 }

maspirin=96.7 g{ m_{\text{aspirin}} = 96.7\, \text{g} }

Conclusion

In this Q&A article, we answered common questions related to the production of aspirin from 77.7 g of C7H6O3C_7H_6O_3. We calculated the number of moles of C7H6O3C_7H_6O_3, the number of moles of C9H8O4C_9H_8O_4, the mass of C9H8O4C_9H_8O_4 produced, and the mass of aspirin produced with 95.0% yield. This article demonstrates the importance of understanding chemical reactions and stoichiometry in chemistry.

References

Glossary

  • Molar mass: The mass of one mole of a substance.
  • Mole: A unit of measurement for the amount of a substance.
  • Stoichiometry: The study of the quantitative relationships between reactants and products in chemical reactions.
  • Yield: The percentage of the desired product obtained from a chemical reaction.