Question

this is the chemical formula for methyl tert - butyl ether (the clean - fuel gasoline additive mtbe):
ch₃oc(ch₃)₃
a chemical engineer has determined by measurements that there are 0.071 moles of hydrogen in a sample of methyl tert - butyl ether. how many moles of oxygen are in the sample?
round your answer to 2 significant digits.
mol

Explanation:

Step1: Determine the ratio of hydrogen to oxygen in MTBE

The chemical formula for MTBE is \(CH_3OC(CH_3)_3\). Expanding it, we get \(C_5H_{12}O\). So, the ratio of \(H\) to \(O\) is \(12:1\).

Step2: Set up a proportion to find moles of oxygen

Let \(x\) be the moles of \(O\). We know \(\frac{\text{moles of }H}{\text{moles of }O}=\frac{12}{1}\). Given moles of \(H = 0.071\) mol. So, \(\frac{0.071}{x}=\frac{12}{1}\).

Solving for \(x\), we get \(x=\frac{0.071}{12}\).

Step3: Calculate the value of \(x\)

\(x=\frac{0.071}{12}\approx0.005917\)

Step4: Round to 2 significant digits

Rounding \(0.005917\) to 2 significant digits gives \(0.0059\approx0.0059\) (but wait, \(0.005917\), the first two non - zero digits are \(5\) and \(9\). Since the next digit \(1<5\), we keep it as \(0.0059\). Wait, no: \(0.071\div12 = 0.005916\cdots\). In terms of significant digits, \(0.071\) has 2 significant digits. So \(0.005916\cdots\approx0.0059\) (but actually, \(0.071\div12=\frac{7.1\times10^{-2}}{1.2\times10^{1}}=\frac{7.1}{1.2}\times10^{-3}\approx5.9\times10^{-3}\))

Answer:

\(0.0059\) mol