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hodyreva [135]
3 years ago
12

How many moles of c9h8o4 are in a 0.600 g tablet of aspirin?

Chemistry
2 answers:
lord [1]3 years ago
4 0

Answer : The moles of C_9H_8O_4 are, 108.0948 mole

Explanation : Given,

Mass of aspirin = 0.600 g

Molar mass of aspirin = 180.158 g/mole

The molecular formula of aspirin is, C_9H_8O_4

Formula used :

\text{Moles of }C_9H_8O_4=\frac{\text{Mass of }C_9H_8O_4}{\text{Molar mass of }C_9H_8O_4}

Now put all the given values in this formula, we get the moles of C_9H_8O_4.

\text{Moles of }C_9H_8O_4=\frac{0.600g}{180.158g/mole}=108.0948mole

Therefore, the moles of C_9H_8O_4 are, 108.0948 mole

larisa86 [58]3 years ago
4 0

\boxed{{\text{0}}{\text{.0033}}\;{\text{mol}}} of {{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}} are present in 0.600 g tablet of aspirin.

Further Explanation:

Mole is a measure of the amount of substance. It is defined as the mass of a substance that has the same number of fundamental units as there are atoms in 12 g of carbon-12. Such fundamental units can be atoms, molecules or formula units.

In step 1, we have to calculate the molar mass of aspirin. Molar mass of aspirin can be calculated by using the equation 1.

Step 1: The molecular formula of aspirin is {{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}}. Its molar mass is calculated as follows:

{\text{Molar mass of }}{{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}}=\left[\begin{gathered}{\text{9}}\left( {{\text{Atomic mass of C}}}\right)+\hfill\\{\text{8}}\left( {{\text{Atomic mass of H}}}\right)+\hfill\\{\text{4}}\left( {{\text{Atomic mass of O}}}\right)\hfill\\\end{gathered}\right]             …… (1)

The atomic mass of C is 12 g.

The atomic mass of H is 1 g.

The atomic mass of O is 16 g.

Substitute these values in equation (1).

\begin{aligned}{\text{Molar mass of }}{{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}}&=\left[{9\left( {{\text{12}}} \right)+8\left({\text{1}}\right)+4\left({{\text{16}}}\right)} \right]\\&=108+8+64\\&= {\mathbf{180}}\;{\mathbf{g/mol}}\\\end{aligned}

In step 2, the molar mass calculated in step 1 is used to calculate the number of moles of aspirin. The number of moles of aspirin is calculated by using equation (2).

Step 2: The formula to calculate the moles of aspirin \left( {{{\mathbf{C}}_{\mathbf{9}}}{{\mathbf{H}}_{\mathbf{8}}}{{\mathbf{O}}_{\mathbf{4}}}}\right) is as follows:

{\text{Moles of }}{{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}}=\dfrac{{{\text{Given mass of }}{{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}}}}{{{\text{Molar mass of }}{{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}}}}               ...... (2)

The given mass of aspirin is 0.600 g.

The molar mass of aspirin is 180 g/mol.

Substitute these values in equation (2)

\begin{aligned}{\text{Moles of }}{{\text{C}}_{\text{9}}}{{\text{H}}_{\text{8}}}{{\text{O}}_{\text{4}}}&=\left({0.600\;{\text{g}}} \right)\left( {\frac{{1\;{\text{mol}}}}{{180\;{\text{g}}}}}\right)\\&={\text{0}}{\text{.003333}}\\&\approx {\mathbf{0}}{\mathbf{.0033}}\;{\mathbf{mol}}\\\end{aligned}

Learn more:

1. Calculate the moles of chlorine: brainly.com/question/3064603

2. What is the number of moles of ions in HCl solution? brainly.com/question/5950133

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Mole concept

Keywords: moles, aspirin, 0.600 g, C9H8O4, molar mass, carbon-12, amount of substance, mass of substance, formula units, 180 g/mol, 0.0033 mol.

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Answer:

a. The limiting reactant is NaHCO_{3}

b. 0.73 g of carbon dioxide are formed.

c. The grams of excess reactant that do not participate in the reaction are 0333 g.

Explanation:

a)

You know the following reaction:

3NaHCO_{3} +H_{3} C_{6} H_{5} O_{7}⇒3CO_{2} +3H_{2} O+Na_{3} C_{6} H_{5} O_{7}

First, you determine the molar mass of each compound. For that you must take into account the atomic mass of each element:

  • Na:  23
  • H: 1
  • C: 12
  • O: 16

To determine the molar mass of each compound, you multiply the most atomic of each element present in the molecule by the sub-index that appears after each number, which indicates the present amount of each element in the compound:

  • NaHCO_{3} :23+1+12+16*3=84 g/mol
  • H_{3} C_{6} HO_{7} :1*3+12*6+1*5+16*7= 192 g/mol
  • CO_{2} :12+16*2= 44 g/mol
  • H_{2} O :1*2+16= 18 g/mol
  • Na_{3} C_{6} H_{5} O_{7} : 23*3+12*6+1*5+16*7= 258 g/mol

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you know that 3 moles of NaHCO_{3} react with 1 mole of H_{3} C_{6} HO_{7}  Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 192 g

You know that in a certain experiment you have 1.40 g of sodium bicarbonate and 1.40 g of citric acid. To determine the limiting reagent apply a rule of three simple as follows:  

If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of sodium bicarbonate react with 1.4 grams of citric acid?

grams of sodium bicarbonate= \frac{1.4 g*252 g}{192 g}

grams of sodium bicarbonate= 1.8375 g

But to perform the experiment you have only 1.4 g of sodium bicarbonate. So <u><em>the limiting reagent is sodium bicarbonate</em></u>.

b)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first.

Now, by stoichiometry of the reaction, you know that 3 moles of NaHCO_{3} react with 3 mole of CO_{2}. Then, taking into account the molar mass of each compound, you can calculate the reacting mass of each compound by stoichiometry:

  • NaHCO_{3} : 252 g
  • H_{3} C_{6} HO_{7} : 132 g

You make a simple rule of three: if 252 g of sodium bicarbonate form 132 g of carbon dioxide per stochetry, how many grams will form 1.4 g of sodium bicarbonate?

grams of carbon dioxide =\frac{1.4 g * 132 g}{252 g}

<u><em>grams of carbon dioxide=  0.73 g</em></u>

<u><em>Then, 0.73 g of carbon dioxide are formed.</em></u>

c)

As mentioned, the limiting reagent is sodium bicarbonate. This means that you should use 1.4 g of sodium bicarbonate for all subsequent calculations, because this compound is the reagent that will be consumed first. This means that citric acid will not react everything, leaving an excess.

To know how much citric acid will react you apply a rule of three, taking into account as in the previous cases the stoichiometry of the reaction: If by stoichiometry 252 g of sodium bicarbonate react with 192 g of citric acid, how many grams of citric acid will they react with 1.4 g of sodium bicarbonate?

grams of citric acid=\frac{1.4 g * 192 g}{252 g}

grams of citric acid= 1.067 g

But you have 1.4 g of citric acid. That means that the grams you have minus the grams that react will be the grams that remain in excess and do not participate in the reaction:

grams of excess reactant=1.4 g - 1.067 g

grams of excess reactant=0.333 g

<em><u>So the grams of excess reactant that do not participate in the reaction are 0333 g.</u></em>

3 0
3 years ago
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