Question

1. azurite is a copper containing mineral that is a mixture of copper(ii) carbonate and copper (ii) hydroxide: cuco₃ₓ - cu(oh)₂ᵧ analysis reveals the following percentages by mass: 55.31% cu, .58% h, and 6.97% c. the balance is oxygen. determine the values of x and y.

Explanation:

Step1: Calculate moles of each element

Assume we have a 100 - g sample of the mineral.

• Moles of \(Cu\): \(n_{Cu}=\frac{55.31\ g}{63.55\ g/mol}\approx0.87\ mol\)
• Moles of \(H\): \(n_{H}=\frac{0.58\ g}{1.008\ g/mol}\approx0.575\ mol\)
• Moles of \(C\): \(n_{C}=\frac{6.97\ g}{12.01\ g/mol}\approx0.58\ mol\)

Step2: Relate moles based on formula

From the formula \([CuCO_{3}]_{x}-[Cu(OH)_{2}]_{y}\), the number of \(Cu\) atoms is \(x + y\), the number of \(C\) atoms is \(x\), and the number of \(H\) atoms is \(2y\).

• Since \(n_{C}=x\approx0.58\ mol\) and \(n_{H} = 2y\approx0.575\ mol\), then \(y=\frac{0.575}{2}\approx0.2875\)
• Also, since \(n_{Cu}=x + y\approx0.87\ mol\), substituting \(x = 0.58\) gives \(y=0.87 - 0.58 = 0.29\)

Rounding to whole numbers (since \(x\) and \(y\) are integers in the formula of a compound), we find that if we multiply by 2 to get rid of the decimals:
Let \(x = 2\) and \(y = 1\)

Check:

• Moles of \(Cu\): \(x + y=2 + 1=3\)
• Moles of \(C\): \(x = 2\)
• Moles of \(H\): \(2y=2\times1 = 2\)

If we calculate the mass percentage for a formula \(2CuCO_{3}-Cu(OH)_{2}\) (molar mass \(M=2\times(63.55 + 12.01+3\times16)+63.55+2\times(16 + 1.008)\)
\(M=2\times(63.55+12.01 + 48)+63.55+2\times(17.008)\)
\(M=2\times123.56+63.55 + 34.016\)
\(M = 247.12+63.55+34.016=344.686\ g/mol\)

Mass of \(Cu\): \(3\times63.55 = 190.65\ g\), mass percentage of \(Cu=\frac{190.65}{344.686}\times100\%\approx55.3\%\)
Mass of \(C\): \(2\times12.01=24.02\ g\), mass percentage of \(C=\frac{24.02}{344.686}\times100\%\approx6.97\%\)
Mass of \(H\): \(2\times2\times1.008 = 4.032\ g\), mass percentage of \(H=\frac{4.032}{344.686}\times100\%\approx1.17\%\) (There was a miscalculation in the initial mole - ratio due to rounding in the first approach. Let's use another method)

Let the formula be \(aCuCO_{3}+bCu(OH)_{2}\)
The molar mass of \(CuCO_{3}=63.55 + 12.01+3\times16=123.56\ g/mol\)
The molar mass of \(Cu(OH)_{2}=63.55+2\times(16 + 1.008)=97.566\ g/mol\)

The molar mass of the compound \(M=a\times123.56 + b\times97.566\)
Mass of \(Cu\): \((a + b)\times63.55\)
Mass of \(C\): \(a\times12.01\)
Mass of \(H\): \(2b\times1.008\)

We know that \(\frac{(a + b)\times63.55}{a\times123.56 + b\times97.566}\times100\%=55.31\%\)
\(\frac{a\times12.01}{a\times123.56 + b\times97.566}\times100\%=6.97\%\)
\(\frac{2b\times1.008}{a\times123.56 + b\times97.566}\times100\%=0.58\%\)

From \(\frac{2b\times1.008}{a\times123.56 + b\times97.566}\times100\%=0.58\%\), we have \(2b\times1.008=0.0058\times(a\times123.56 + b\times97.566)\)
\(2.016b=0.7166a+0.5659b\)
\(2.016b- 0.5659b=0.7166a\)
\(1.4501b=0.7166a\), so \(a = 2b\)

Substitute \(a = 2b\) into \(\frac{(a + b)\times63.55}{a\times123.56 + b\times97.566}\times100\%=55.31\%\)
\(\frac{(2b + b)\times63.55}{2b\times123.56 + b\times97.566}\times100\%=55.31\%\)
\(\frac{3b\times63.55}{b\times(247.12+97.566)}\times100\%=55.31\%\)
\(\frac{190.65}{344.686}\times100\%=55.31\%\) (verified)

Answer:

\(x = 2,y = 1\)