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4 years ago

Assuming the compound is dissolved in water, what is the formula for phosphorous acid?

Chemistry

2 answers:

LekaFEV [45]4 years ago

8 0

Answer is: the formula for phosphorous acid is H₃PO₃.

Phosphorous acidis diprotic (lost two protons).

Dissociation of phosphorous acid in water:

H₃PO₃(aq) → 2H⁺(aq) + PHO₃²⁻(aq).

Solubility in water of this acid is 310 g/100 mL.

Other oxyacids of phosphorus are phosphoric acid (H₃PO₄) and hypophosphorous acid (H₃PO₂).

lozanna [386]4 years ago

7 0

H3PO3 because this compound dissolved in water will give you that formula

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Answer: The empirical formula for the given compound is $C_{4}H_{10}O$

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The chemical equation for the combustion of ether follows:

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

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For calculating the mass of hydrogen:

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

So, in 2.128 g of water, $\frac{2}{18}\times 2.128=0.236g$ of hydrogen will be contained.

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.134g}{12g/mole}=0.0945moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.236g}{1g/mole}=0.236moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.381g}{16g/mole}=0.0238moles$

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 0.0238 moles.

For Carbon = $\frac{0.0945}{0.0238}=3.97\approx 4$

For Hydrogen = $\frac{0.236}{0.0238}=9.91\approx 10$

For Oxygen = $\frac{0.0238}{0.0238}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 10 : 1

Hence, the empirical formula for the given compound is $C_{4}H_{10}O$

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