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Fittoniya [83]
3 years ago
12

Determine the empirical formulas for compounds with the following percent compositions:

Chemistry
1 answer:
Elis [28]3 years ago
3 0

<u>Answer:</u>

<u>For a:</u> The empirical formula of the compound is P_2O_5

<u>For b:</u> The empirical formula of the compound is KH_2PO_4

<u>Explanation:</u>

  • <u>For a:</u>

We are given:

Percentage of P = 43.6 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of P = 43.6 g

Mass of O = 56.4 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{43.6g}{31g/mole}=1.406moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{56.4g}{16g/mole}=3.525moles

  • <u>Step 2:</u> 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 1.406 moles.

For Phosphorus = \frac{1.406}{1.406}=1

For Oxygen = \frac{3.525}{1.406}=2.5

Converting the moles in whole number ratio by multiplying it by '2', we get:

For Phosphorus = 1\times 2=2

For Oxygen = 2.5\times 2=5

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of P : O = 2 : 5

Hence, the empirical formula for the given compound is P_2O_5

  • <u>For b:</u>

We are given:

Percentage of K = 28.7 %

Percentage of H = 1.5 %

Percentage of P = 22.8 %

Percentage of O = 56.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of K = 28.7 g

Mass of H = 1.5 g

Mass of P = 43.6 g

Mass of O = 56.4 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Potassium =\frac{\text{Given mass of Potassium}}{\text{Molar mass of Potassium}}=\frac{28.7g}{39g/mole}=0.736moles

Moles of Hydrogen =\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of hydrogen}}=\frac{1.5g}{1g/mole}=1.5moles

Moles of Phosphorus =\frac{\text{Given mass of Phosphorus}}{\text{Molar mass of Phosphorus}}=\frac{22.8g}{31g/mole}=0.735moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{47g}{16g/mole}=2.9375moles

  • <u>Step 2:</u> 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 0.735 moles.

For Potassium = \frac{0.736}{0.735}=1

For Hydrogen = \frac{1.5}{0.735}=2.04\approx 2

For Phosphorus = \frac{0.735}{0.735}=1

For Oxygen = \frac{2.9375}{0.735}=3.99\approx 4

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of K : H : P : O = 1 : 2 : 1 : 4

Hence, the empirical formula for the given compound is KH_2PO_4

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Answer:

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Explanation:

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Moles of CO₃²⁻ = K₂CO₃ are:

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0.009570 moles ₓ (84.007g / mol) =

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