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dedylja [7]
4 years ago
13

An unknown compound has the empirical formula CH2O and a molecular mass of 180 amu. What is the molecular formula of the compoun

d
Chemistry
1 answer:
Sav [38]4 years ago
6 0
The empirical formula gives the simplest ratio of whole numbers of components in the compound. Molecular formula gives the actual composition of elements in the compound.
empirical formula - CH₂O
Mass of one empirical unit = 12 + (1x2) + 16 = 30 a.m.u
We have to next calculate how many empirical units make up the molecular formula. For that we have to divide molecular mass by mass of one empirical unit.
Number of empirical units = 180 a.m.u / 30 = 6
There are 6 empirical units 
Therefore molecular formula = 6 x(CH₂O) = C₆H₁₂O₆
Molecular formula = C₆H₁₂O₆
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Explanation:

According to the law of conservation of mass, mass can neither be created nor destroyed but it can simply be transformed from one form to another.

For example, Na^{+} + Cl^{-} \rightarrow NaCl

Mass of Na = 23 g/mol

Mass of Cl = 35.5 g/mol

Sum of mass of reactants = mass of Na + mass of Cl

= 23 + 35.5 g/mol

= 58.5 g/mol

Mass of product formed is as follows.

Mass of NaCl = mass of Na + mass of Cl

= (23 g/mol + 35.5) g/mol

= 58.5 g/mol

As mass reacted is equal to the amount of mass formed. This shows that mass is conserved.

As a result, law of conservation of mass is obeyed.

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6. What is the molecular formula for this compound? The compound's empirical formula and
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Explanation:

A compound's empirical formula tells you the smallest whole number ratio ,The molar mass tells you what the total mass of one mole

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The reaction of ethane gas (C2H6) with chlorine gas produces C2H5Cl as its main product (along with HCl). In addition, the react
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Answer:

The percent yield of  chloro-ethane in the reaction is 82.98%.

Explanation:

C_2H_6+Cl_2\rightarrow C_2H_5Cl+HCl

Moles of ethane = \frac{300.0 g}{30 g/mol}=10 mol

Moles of chlorine gases =\frac{650.0 g}{71 .0 g/mol}=9.1549 mol

As we can see that 1 mol of ethane react with 1 mole of chlorine gas.the 10 moles will require 10 mole of chlorine gas, but only 9.1549 moles of chlorine gas is present.

This means that chlorine gas is in limiting amount and amount of formation of chloro-ethane will depend upon amount of chlorine gas.

According to reaction , 1 mol of chloro ethane gives 1 mol of chloro-ethane.

Then 9.1549 moles of chlorien gas will give:

\frac{1}{1}\times 9.1549 mol=9.1549 mol of chloro-ethane

Mass of 9.1549 moles of chloro-ethane:

9.1549 mol × 64.5 g/mol = 590.4910 g

Theoretical yield of  chloro-ethane: 590.4910 g

Given experimental yield of chloro-ethane: 490.0 g

\% Yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

\%Yield (C_2H_5Cl)=\frac{490.0 g}{590.4910 g}\times 100=82.98\%

The percent yield of  chloro-ethane in the reaction is 82.98%.

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