Question

A 59.4-mg sample of the compound x4o6 contains 14.4 mg of oxygen atoms. What is the atomic mass of element x?

Answer 1

`m_X=75.1g/mol`

Step-by-step explanation:

Hello,

One can solve this problem by using percent compositions, therefore, the first step is to compute
`X`'s percent as shown below:

`\% X=(59.4mg-14.4mg)/(59.4mg)*100 \% = 75.8\%`

Now, we define the formula for the percent composition of
`X` in
`X_4O_6` as follows:

`\% X=(4*m_X)/(M_(X_4O_6))`

Whereas
`M_(X_4O_6)` is the molar mass of the given compound and
`m_X` the atomic mass of
`X`, thus, we obtain:

`75.8\% =(4*m_X)/(16*6+4*m_X)`

Solving for the atomic mass:

`75.8\% (16*6+4*m_X)=4*m_X\\72.8+3.032m_X=4*m_X\\m_X=(72.8)/(4-3.032) \\m_X=75.1g/mol`

The element having such atomic mass is arsenic (74.9g/mol).

Best regards.

Answer 2

Answer is: the atomic mass of element X is 75.

n(O) = m(O) : M(O).

n(O) = 14.4 mg ÷ 16 g/mol.

n(O) = 0.9 mmol; amount of substance.

From molecular formula of compound: n(O) : n(X) = 6 : 4 (3 : 2).

n(X) = 0.6 mmol.

m(X) = 59.4 mg - 14.4 mg.

m(X) = 45 mg; mass of element X.

M(X) = 45 mg ÷ 0.6 mmol.

M(X) = 75 g/mol.