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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.

sebassandin

Answer:

[tex]m_X=75.1g/mol[/tex]

Explanation:

Hello,

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

[tex]\% X=\frac{59.4mg-14.4mg}{59.4mg}*100 \% = 75.8\%[/tex]

Now, we define the formula for the percent composition of [tex]X[/tex] in [tex]X_4O_6[/tex] as follows:

[tex]\% X=\frac{4*m_X}{M_{X_4O_6}}[/tex]

Whereas [tex]M_{X_4O_6}[/tex] is the molar mass of the given compound and [tex]m_X[/tex] the atomic mass of [tex]X[/tex], thus, we obtain:

[tex]75.8\% =\frac{4*m_X}{16*6+4*m_X}[/tex]

Solving for the atomic mass:

[tex]75.8\% (16*6+4*m_X)=4*m_X\\72.8+3.032m_X=4*m_X\\m_X=\frac{72.8}{4-3.032} \\m_X=75.1g/mol[/tex]

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

Best regards.

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