Question

a sample of copper carbonate was found to contain 15.9 g of copper. how many grams of copper carbonate were in the sample? use the ratio of copper to oxygen to carbon. a. 3 grams b. 5.3 grams c. 30.9 grams d. 47.7 grams

Explanation:

Step1: Determine the ratio of elements in copper carbonate

The formula of copper carbonate is \(CuCO_3\). The atomic masses: \(Cu = 63.5\space g/mol\), \(C=12\space g/mol\), \(O = 16\space g/mol\). The molar mass of \(CuCO_3=63.5 + 12+3\times16=63.5 + 12 + 48=123.5\space g/mol\). The mass of \(Cu\) in \(CuCO_3\) is \(63.5\space g/mol\).

Step2: Set up a proportion

Let \(x\) be the mass of \(CuCO_3\). We know that the ratio of \(Cu\) mass to \(CuCO_3\) mass is \(\frac{63.5}{123.5}\). Given the mass of \(Cu\) is \(15.9\space g\). So \(\frac{63.5}{123.5}=\frac{15.9}{x}\). Cross - multiply: \(63.5x=15.9\times123.5\). Then \(x=\frac{15.9\times123.5}{63.5}\).

Step3: Calculate the value of \(x\)

\(15.9\times123.5 = 15.9\times(120 + 3.5)=15.9\times120+15.9\times3.5=1908+55.65 = 1963.65\). Then \(x=\frac{1963.65}{63.5}\approx30.9\space g\)

Answer:

C. 30.9 grams