Question

two oxides of metal contains 63.2% and 69.62% of the metal.the specific heat of the metal is 0.117.what is the formula of the two oxides?

Answer 1

Vanadium is the metal with a specific heat of 0.117 J/goC

 

 

 

V = 51

 

First oxide

 

Moles of V = 63.2/51 = 1.24

Moles of O = 36.8/16 = 2.3

 

Mole ratio

V  = 1.24/1.24 = 1

O = 2.3/1.24 = 2

 

Formula: VO2

 

 

 

Second oxide

 

Moles of V= 69.62/51 = 1.365

Moles of O = 30.38/16 = 1.9

 

Mole ratio

V = 1.365/1.365 = 1

O = 1.9/1.365 = 1.4

 

1 : 1.4 = 5:7

 

Formula: V5O7

 

 

Answer 2

The formula of metal oxides are $\mathrm { VO } _ { 2 } \quad and \quad \mathrm {V}_{5} {O}_{7}$

Given:

Percentage of Metal V= 63.2%

Percentage of Second Metal  V = 69.62%

To find:

The formula of two metal oxides

Solution:

Metal oxide 1:

The Vanadium has specific heat is 0.117 $\mathrm { J } / \mathrm { g } \mathrm { C } ^ { 0 }$

Percentage of Metal V= 63.2%

$Moles \quad of \quad V =\frac {\text (Given \quad mass)}{\text (Molar \quad mass)} = \frac{63.2}{51}=1.24$

Percentage of Oxygen = 100 – 63.2 = 36.8

$Moles \quad of \quad O = \frac {36.8}{16}=2.3$

Mole ratio:

$For \quad V = \frac {1.24}{1.24}=1$

$For \quad O= \frac {2.3}{1.24}= 2$

Therefore, formula of metal oxide 1 is $\mathrm { VO } _ { 2 }$

Metal oxide 2:

Metal  V = 69.62

Percentage of oxygen = 100 – 69.62 = 30.38

$Moles \quad of \quad V =\frac {69.62}{51}=1.36$

$Moles \quad of \quad O =\frac {30.38}{16}=1.9$

Mole ratio:

$For \quad V =\frac {1.36}{1.36}= 1$

$For \quad O=\frac {1.9}{1.365}=1.4$

$1 : 1.4 \approx 5 : 7$

Therefore, molecular formula of second metal oxide is$\mathrm { V } _ { 5 } \quad and \quad \mathrm { O } _ { 7 }$