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schepotkina [342]
3 years ago
13

The empirical formula of an organic compound is C2H4O. The molecular mass of the compound is 176g/mol.

Chemistry
1 answer:
Brrunno [24]3 years ago
6 0

Answer:

The molecular formula of the compound is C_{8}H_{16}O_{4}. The molecular formula is obtained by the following expression shown below

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is  C_{2}H_{4}O

Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.

Empirical formula mass of the compound = \left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}

n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4

\textrm{Molecular formula }= n\times \textrm{Empirical formula}

Molecular formula = 4 \times C_{2}H_{4}O

Molecular formula is C_{8}H_{16}O_{4}

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Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation,
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Answer : The full balanced ionic equation will be,

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, K^+,NO_3^-

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)

The ionic equation in separated aqueous solution will be,

2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)

In this equation, K^+\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)

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