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Dmitrij [34]
3 years ago
10

Manganese sulfate forms a pale pink hydrate with the formula MnSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high enough

temperature, H 2 O ( g ) can be driven off, leaving the grey‑white anhydrous salt MnSO 4 ( s ) . A 16.260 g sample of the hydrate was heated to 300 ∘ C . The resulting MnSO 4 ( s ) had a mass of 14.527 g . Calculate the value of n in MnSO 4 ⋅ n H 2 O ( s ) .
Chemistry
1 answer:
Mama L [17]3 years ago
8 0

Answer:

Value of n in MnSO₄.nH₂O is one.

Explanation:

The n represents the number of moles of water attached to the formula unit manganese sulfate. These moles (n) can be determined by taking the ratio of the moles of anhydrous salt and the moles of water. The moles of water can be determined by taking the difference of final and initial mass of the salt. This difference is equal to the mass of the water, mathematically it can be represented as,

Mass of H₂O = initial mass of the salt (g) - final mass of the salt (g)

Mass of H₂O = 16.260 g - 14.527 g

Mass of H₂O = 1.733 g

moles of H₂O = (1.733 g) ÷ (18.015 g/mole)

moles of H₂O =  0.0962

For the moles of anhydrous salt:

moles of MnSO₄ = mass of MnSO₄ ÷ molar mass of MnSO₄

moles of MnSO₄ = 14.5277 ÷ 151.001

moles of MnSO₄= 0.0962

Now for n:

n = moles of water ÷ moles of MnSO₄

n = 0.0962 ÷ 0.0962

n = 1

The above calculations show that one mole of H₂O is attached to the  one formula unit of MnSO₄

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<u>Answer:</u> The empirical formula for the given compound is C_3H_6O

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

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We know that:

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  • <u>For calculating the mass of carbon:</u>

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So, in 0.00632 g of carbon dioxide, \frac{12}{44}\times 0.00632=0.00172g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

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

So, in 0.00258 g of water, \frac{2}{18}\times 0.00258=0.000286g of hydrogen will be contained.

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

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

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles

  • <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 4.83\times 10^{-5}mol

For Carbon = \frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3

For Hydrogen  = \frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6

For Oxygen  = \frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1

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The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is C_3H_{6}O_1=C_3H_6O

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