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lianna [129]
3 years ago
5

Define molecular formula and empirical formula. What are the similarities and differences between the empirical formula and mole

cular formula of a compound?
Chemistry
1 answer:
irga5000 [103]3 years ago
6 0

<u>Explanation:</u>

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.  

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.  

In both the formulas, the nature of atoms remains the same but the number differs.

For Example: The molecular formula of oxalic acid is C_2H_2H_4 but the empirical formula of oxalic acid is CHO_2

To calculate the molecular formula, we need to find the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

The empirical mass can be calculated from empirical formula and molar mass must be known.

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<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

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Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

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According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

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Put the value of [A'] , [B'] and r' in the above equation:

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Divide equation (1) by (2) we , get

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2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

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B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

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