Question

a 53.81 gram sample of copper is heated in the presence of excess oxygen. a metal oxide is formed with a mass of 60.59 g. determine the empirical formula of the metal oxide. enter the elements in the order cu, o empirical formula =

Explanation:

Step1: Calculate moles of Cu

The molar - mass of Cu is approximately 63.55 g/mol. The mass of Cu is 53.81 g. Using the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass, and $M$ is the molar - mass. So, $n_{Cu}=\frac{53.81\ g}{63.55\ g/mol}\approx0.847\ mol$.

Step2: Calculate moles of O

The mass of the metal oxide is 60.59 g. The mass of O is $m_O = 60.59\ g−53.81\ g = 6.78\ g$. The molar - mass of O is approximately 16.00 g/mol. Using the formula $n=\frac{m}{M}$, we get $n_O=\frac{6.78\ g}{16.00\ g/mol}\approx0.424\ mol$.

Step3: Find the mole - ratio

Divide the number of moles of each element by the smaller number of moles. For Cu: $\frac{n_{Cu}}{n_O}=\frac{0.847\ mol}{0.424\ mol}\approx2$. For O: $\frac{n_O}{n_O} = 1$.

Answer:

$Cu_2O$