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Norma-Jean [14]
2 years ago
8

pyrite is a mineral composed of 46.5 mass % iron and 53.5 mass % sulfur. determine the empirical formula for calcocite.

Chemistry
1 answer:
padilas [110]2 years ago
3 0

The empirical formula for pyrite is FeS2.

HOW TO CALCULATE EMPIRICAL FORMULA:

  • The empirical formula represents the simplest whole number ratio of constituents element of a compound. The empirical formula of pyrite can be calculated as follows:

46.5 mass % Fe = 46.5g of Fe

53.5 mass % S = 53.5g of S

  • Next, we divide each element's mass value by its molar mass

Fe = 46.5g ÷ 56g/mol = 0.83mol

S = 53.5g ÷ 32g/mol = 1.67mol

  • Next, we divide each mole value by the smallest (0.83mol)

Fe = 0.83mol ÷ 0.83 = 1

S = 1.67mol ÷ 0.83 = 2.014

Approximately, the ratio of Fe to S is 1:2. Therefore, the empirical formula of pyrite is FeS2.

Learn more at: brainly.com/question/14044066?referrer=searchResults

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In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, \frac{12}{44}\times 21363=5826.27g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, \frac{2}{18}\times 6125=680.55 of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = \frac{485.52}{97.73}=4.96\approx 5

For Hydrogen  = \frac{680.55}{97.73}=6.96\approx 7

For Nitrogen = \frac{97.73}{97.73}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is C_5H_7N_1=C_5H_7N

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