Question

If the mass of a proton is doubled and that of neutron is halved, the molecular weight of $CO_2$, consisting only $C^{12}$ and $O^{16}$ atoms, will

• A. not change
• B. increases by 25%
• C. decrease by 25%
• D. increase by 50%

Answer 1

Answer: (B)

increases by 25%

Answer 2

Let $m_p$ be the original mass of a proton and $m_n$ be the original mass of a neutron. A $C^{12}$ atom has 6 protons and 6 neutrons. An $O^{16}$ atom has 8 protons and 8 neutrons. In a $CO_2$ molecule, there is one $C^{12}$ atom and two $O^{16}$ atoms. Total protons in $CO_2 = 6 + 2 \times 8 = 22$. Total neutrons in $CO_2 = 6 + 2 \times 8 = 22$. Initial molecular weight ($MW_{initial}$) = $22 m_p + 22 m_n$.

New proton mass $m_p' = 2m_p$. New neutron mass $m_n' = m_n/2$. New molecular weight ($MW_{new}$) = $22 m_p' + 22 m_n'$ = $22(2m_p) + 22(m_n/2) = 44m_p + 11m_n$.

Percentage change = $\frac{MW_{new} - MW_{initial}}{MW_{initial}} \times 100\% = \frac{(44m_p + 11m_n) - (22m_p + 22m_n)}{22m_p + 22m_n} \times 100\% = \frac{22m_p - 11m_n}{22m_p + 22m_n} \times 100\%$. Assuming $m_p \approx m_n = m$, Percentage change = $\frac{22m - 11m}{22m + 22m} \times 100\% = \frac{11m}{44m} \times 100\% = \frac{1}{4} \times 100\% = 25\%$. Since $MW_{new} > MW_{initial}$, it is an increase.