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garik1379 [7]
2 years ago
14

The elemental mass percent composition of ascorbic acid (vitamin C ) is 40.92% C , 4.58% H , and 54.50% O . Determine the empiri

cal formula of ascorbic acid.
Chemistry
1 answer:
Lady bird [3.3K]2 years ago
7 0

The empirical formula of ascorbic acid is C6H8O6.

<h3>Empirical formula</h3>

To calculate the empirical mass of a compound from the mass percentage of each element, the value of the<u> molar mass </u>of each element is used, which in this case corresponds to:

                                              MM_C= 12g/mol\\MM_H=1g/mol\\MM_O=16g/mol

From this, we consider that the compound has 100 grams, so the number of moles of each element will be equal to:

                                                   C = \frac{40.92}{12} = 3.41 \\H = \frac{4.58}{1}=4.58\\ O =\frac{54.50}{16}=3.41

Now, divide all the values ​​found by the <u>smallest value</u>, to find the amount present in each element:

                                            C = \frac{3.41}{3.41} = 1\\ H =  \frac{4.58}{3.41} = 1.34\\O =  \frac{3.41}{3.41} = 1

As you can see, the value of moles of hydrogen resulted in a decimal number, so it is necessary to multiply all values ​​​​by a number in common until the three meet as integers:

                                          C = 1 \times 6 = 6\\H = 1.34 \times 6 = 8\\O = 6 \times 6 = 6

So, the empirical formula of ascorbic acid is C6H8O6.

Learn more about empirical and molecular formula in: brainly.com/question/11588623

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Many computer chips are manufactured from silicon, which occurs in nature as SiO2. When SiO2 is heated to melting, it reacts wit
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<u>Answer:</u>

<u>(a):</u> The theoretical yield of silicon is 72.33 kg.

<u>(b):</u> The percent yield of the reaction is 91.25 %.

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       .....(1)

<u>For </u>SiO_2<u>:</u>

Given mass = 155.0 kg = 155000 g          (Conversion factor: 1 kg = 1000 g)

Molar mass = 60 g/mol

Putting values in equation 1:

\text{Moles of }SiO_2=\frac{155000g}{60g/mol}=2583.3mol

<u>For carbon:</u>

Given mass = 78.2 kg = 78200 g

Molar mass = 12 g/mol

Putting values in equation 1:

\text{Moles of carbon}=\frac{78200g}{12g/mol}=6516.67mol

The chemical equation for the reaction of silicon dioxide and carbon follows:

SiO_2+2C\rightarrow Si+2CO

By stoichiometry of the reaction:

1 mole of SiO_2 reacts with 2 moles of carbon

So, 2583.3 moles of SiO_2 will react with = \frac{2}{1}\times 2583.3=5166.4mol of carbon

As the given amount of carbon is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, SiO_2 is considered a limiting reagent because it limits the formation of the product.

  • <u>For (a):</u>

By stoichiometry of the reaction:

1 mole of SiO_2 produces 1 mole of silicon

So, 2583.3 moles of SiO_2 will produce = \frac{1}{1}\times 2583.3=2583.3mol of silicon

Since the molar mass of silicon = 28 g/mol

Putting values in equation 1:

\text{Mass of Si}=2583.3mol\times 28g/mol=72332.4g=72.33 kg

Hence, the theoretical yield of silicon is 72.33 kg.

  • <u>For (b):</u>

The percent yield of a reaction is calculated by using an equation:

\% \text{yield}=\frac{\text{Measured value}}{\text{Theoretical value}}\times 100              ......(2)

Given values:

Measured value of silicon = 66.0 kg

Theoretical value of silicon  = 72.33 kg

Putting values in equation 1:

\% \text{yield}=\frac{66.0kg}{72.33kg}\times 100\\\\\% \text{yield}=91.25 \%

Hence, the percent yield of the reaction is 91.25 %.

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