A metallic element M forms two distinct oxides. Quantitative analysis shows that the first… — JAMB UTME

A metallic element $M$ forms two distinct oxides. Quantitative analysis shows that the first oxide contains $20.0\%$ oxygen by mass and has the empirical formula $MO$ . The second oxide contains $11.1\%$ oxygen by mass. Based on the Law of Multiple Proportions, what is the empirical formula of the second oxide?

$M_2O$ Cevap
B
$MO_2$
C
$M_2O_3$
D
$M_3O_4$

Cevap

M_2O

According to the Law of Multiple Proportions, when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. In the first oxide (

MO

20.0\text{ g}

of oxygen combines with

80.0\text{ g}

of metal

M

, giving a ratio of

4.0\text{ g } M

per

1.0\text{ g } O

. In the second oxide,

11.1\text{ g}

of oxygen combines with

88.9\text{ g}

M

, giving a ratio of

8.0\text{ g } M

per

1.0\text{ g } O

. Comparing the mass of metal combining with

1.0\text{ g}

of oxygen gives

4.0 : 8.0 = 1 : 2

. Since the first oxide has

1

atom of

M

per atom of

O

, the second oxide contains

2

atoms of

M

per atom of

O

, yielding the formula

M_2O

Adım Adım Çözüm

Calculate the mass ratio of metal

M

to oxygen

O

in the first oxide

100\text{ g}

of the first oxide, mass of

O = 20.0\text{ g}

and mass of

M = 80.0\text{ g}

. Mass ratio

M : O = \frac{80.0\text{ g } M}{20.0\text{ g } O} = 4.0\text{ g } M / \text{g } O

Determining the mass of metal that combines with a fixed unit mass (

1.0\text{ g}

) of oxygen in the first compound.

Calculate the mass ratio of metal

M

to oxygen

O

in the second oxide

100\text{ g}

of the second oxide, mass of

O = 11.1\text{ g}

and mass of

M = 88.9\text{ g}

. Mass ratio

M : O = \frac{88.9\text{ g } M}{11.1\text{ g } O} = 8.01\text{ g } M / \text{g } O \approx 8.0\text{ g } M / \text{g } O

Determining the mass of metal that combines with a fixed unit mass (

1.0\text{ g}

) of oxygen in the second compound.

Apply the Law of Multiple Proportions to compare the masses of metal

M

combining with fixed oxygen

\text{Ratio of masses of } M = 4.0 : 8.0 = 1 : 2.

The Law of Multiple Proportions states that the masses of one element combining with a fixed mass of another are in simple whole-number ratios.

Deduce the empirical formula of the second oxide

Since the first oxide (

MO

) has

1

atom of

M

per atom of

O

, an oxide with twice the mass of

M

per atom of

O

must have

2

atoms of

M

per atom of

O

, giving the empirical formula

M_2O

Relating the atomic ratio of the second oxide directly to the known formula of the first oxide.

Anahtar Kavram

Law of Multiple Proportions

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru