Question

vanadium has an atomic mass of 50.9415 amu. it has two common isotopes.one isotopes has a mass of 50.9440 amu and a relative abundance of 99.75% .

Answer 1

Explanation:

Average atomic mass of the  vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu

Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025

Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

m × abundance of isotope(I) + m' × abundance of isotope (II)

50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025

m'= 49.944 amu

The atomic mass of isotope (II) of vanadium is 49.944 amu.

Answer 2

Isotope simply means one of the forms of a chemical element. The mass of the second isotope of vanadium is 49.944 amu

Given that:

$A = 50.9415amu$ --- atomic mass of vanadium

$m_1 = 50.9440amu$ ---- the mass of the first isotope

$R_1 = 99.75\%$ ---- the relative abundance of the first isotope

The following equation can be used to calculate the mass of the second isotope.

$A = m_1 \times R_1 + m_2 \times R_2$

Where:

$R_2 =1 - R_1$

$R_2 =1 - 99.75\%$

$R_2 =0.0025$

So, we have:

$A = m_1 \times R_1 + m_2 \times R_2$

$50.9415 =50.9440 \times 99.75\% + m_2 \times 0.0025$

$50.9415 =50.81664 + m_2 \times 0.0025$

Collect like terms

$50.9415 -50.81664 = m_2 \times 0.0025$

$0.12486= m_2 \times 0.0025$

Solve for $m_2$

$m_2 =0.12486 \div 0.0025$

$m_2 =49.944$

Hence, the mass of the second isotope is 49.944 amu

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