Question

What is the mass, in grams, of 1.30×1021 molecules of aspirin, C9H8O4?

Answer 1

0.39 grams

Further explanation

Given that:

1.30 x 10²¹ molecules of aspirin C₉H₈O₄

Question:

The mass, in grams

The Process:

Firstly, we prepare the molar mass of aspirin C₉H₈O₄.

Relative atomic mass (in rounding to integers):

• C = 12 g/moles
• H = 1 g/moles
• O = 16 g/moles

In aspirin, there are 9 carbon atoms, 8 hydrogen atoms, and 4 oxygen atoms. Therefore the molar mass is given by

$\boxed{(12 \times 9) + (1 \times 8) + (16 \times 4) = 180 \ g/moles}$

Secondly, we calculate the moles of 1.30 x 10²¹ molecules of aspirin.

Remember this, one mole contains 6.02 x 10²³ molecules, which is called The Avogadro's number.

Let's convert 1.30 x 10²¹ molecules to the number of moles.

$\boxed{ \ The \ moles = 1.31 \times 10^{21} \ molecules \times \frac{1 \ mole}{6.02 \times 10^{23} \ molecules} \ }$

We get $\boxed{ \ 2.176 \times 10^{-3} \ moles \ }$

Finally, we calculate the mass in grams.

$\boxed{ \ The \ mass = 2.176 \times 10^{-3} \ moles \times 180 \ \frac{g}{moles} \ }$

Thus the mass of 1.30 x 10²¹ molecules of aspirin C₉H₈O₄ is 0.39 grams.

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Quick Steps

$\boxed{\boxed{ \ The \ mass = 1.30 \times 10^{21} \ molecules \times \frac{1 \ mole}{6.02 \times 10^{23} \ molecules} \times \frac{180 \ grams}{1 \ mole} \ }}$

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Keywords: the mass, in grams, molecules, avogadro's numbers, moles, molar mass, relative atomic mass

Answer 2

The mass of  $1.30 \times {\text{1}}{{\text{0}}^{{\text{21}}}}\;{\text{molecules}}$ of aspirin is $\boxed{0.389{\text{ g}}}$.

Further Explanation:

Avogadro’s number is a mathematical number that determines the number of atoms or molecules in one mole of the substance. The value of Avogadro’s number is ${\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{units}}$. These units can either be atoms or molecules.

There are ${\text{6}}{\text{.022}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}$ in one mole of aspirin. Therefore the number of moles in $1.30 \times {\text{1}}{{\text{0}}^{{\text{21}}}}\;{\text{molecules}}$ of aspirin can be calculated as follows:

 $\begin{aligned}{\text{Moles of aspirin}} &= \left( {1.30 \times {\text{1}}{{\text{0}}^{{\text{21}}}}\;{\text{molecules}}} \right)\left( {\frac{{1{\text{ mol}}}}{{6.022 \times {\text{1}}{{\text{0}}^{{\text{23}}}}\;{\text{molecules}}}}} \right)\\&= 2.159 \times {10^{ - 3}}{\text{ mol}}\\\end{aligned}$

The formula to calculate the moles of aspirin is as follows:

${\text{Moles of aspirin}} = \dfrac{{{\text{Mass of aspirin}}}}{{{\text{Molar mass of aspirin}}}}$          …… (1)                                                  

Rearrange equation (1) for the mass of aspirin.

${\text{Mass of aspirin}} = \left( {{\text{Moles of aspirin}}} \right)\left( {{\text{Molar mass of aspirin}}} \right)$     …… (2)                            

The molar mass of aspirin $\left( {{{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}}} \right)$ can be calculated as follows:

  ${\text{Molar mass of }}{{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}} &= \left[ \begin{aligned}9\left({{\text{Atomic mass of C}}} \right)+ \hfill\\8\left( {{\text{Atomic mass of H}}} \right) +  \hfill\\4\left({{\text{Atomic mass of O}}} \right) \hfill\\\end{aligned}  \right]$                                             …… (3)

The atomic mass of C is 12.01 g.

The atomic mass of H is 1.008 g.

The atomic mass of O is 15.99 g.

Substitute these values in equation (3).

 $\begin{aligned}{\text{Molar mass of }}{{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}} &= \left[ {9\left( {{\text{12}}{\text{.01 g}}} \right) + 8\left( {{\text{1}}{\text{.008 g}}} \right) + 4\left( {{\text{15}}{\text{.99 g}}} \right)} \right] \\&=\left[ {108.09{\text{ g}} + 8.064{\text{ g}} + 63.96{\text{ g}}} \right]\\&= 180.114{\text{ g/mol}}\\\end{aligned}$

The number moles of aspirin is $2.159 \times {10^{ - 3}}{\text{ mol}}$.

The molar mass of aspirin is 180.114 g/mol.

Substitute these values in equation (2).

 $\begin{aligned}{\text{Mass of aspirin}} &= \left( {2.159 \times {{10}^{ - 3}}{\text{ mol}}} \right)\left( {\frac{{{\text{180}}{\text{.114 g}}}}{{1{\text{ mol}}}}} \right)\\&= {\text{0}}{\text{.38887 g}}\\&\approx {\text{0}}{\text{.389 g}}\\\end{aligned}$

Therefore the mass of aspirin is 0.389 g.

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Answer details:

Grade: Senior School

Chapter: Mole concept

Subject: Chemistry

Keywords: aspirin, molar mass, atomic mass, C, H, O, C9H8O4, 0.389 g, mass of aspirin, 180.114 g/mol, 12.01 g, 1.008 g, 15.99 g, moles, molecules, $1.30*10^21$ molecules, $6.022*10^23$ molecules, atoms, Avogadro’s number.