The decomposition of n-butyl ether at 504°C is first order with half-life of 1670 sec. What fraction of initial amount of n-butyl ether remain after 4710 second?

For a first-order reaction, the half-life ($t_{1/2}$) is related to the rate constant ($k$) by the formula:

$$t_{1/2} = \frac{\ln(2)}{k}$$

Rearranging to solve for $k$:

$$k = \frac{\ln(2)}{t_{1/2}}$$

Substitute the given half-life $t_{1/2} = 1670 \text{ s}$:

$$k = \frac{0.6931}{1670 \text{ s}}$$ $$k \approx 0.0004150 \text{ s}^{-1}$$

The integrated rate law for a first-order reaction is:

$$\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$$

Where $[A]_t$ is the concentration at time $t$, and $[A]_0$ is the initial concentration. To find the fraction remaining, $\frac{[A]_t}{[A]_0}$, we exponentiate both sides:

$$\frac{[A]_t}{[A]_0} = e^{-kt}$$

Substitute the calculated $k$ and the given time $t = 4710 \text{ s}$:

$$\frac{[A]_t}{[A]_0} = e^{-(0.0004150 \text{ s}^{-1} \times 4710 \text{ s})}$$ $$\frac{[A]_t}{[A]_0} = e^{-1.9545}$$ $$\frac{[A]_t}{[A]_0} \approx 0.1416$$

The fraction of diethyl ether remaining after $4710 \text{ s}$ is approximately $\boxed{\text{0.1416}}$.