Stewart Dent is asked to measure the density of a piece of iron. He measures the mass of the iron and records a value of 97.4 g . He measures the volume of water in a graduated cylinder as 50.4 mL ; after submerging the iron, the volume is 63.3 mL . If the actual density of iron is \( 7.874 \mathrm{~g} / \mathrm{mL} \), what is Stewart's \%-error?

To find Stewart's percent error in the density measurement, we first need to calculate the density of the piece of iron based on his measurements. The density (\(D\)) is given by the formula: \[ D = \frac{\text{mass}}{\text{volume}} \] The mass of the iron is \(97.4 \, \text{g}\), and the volume can be calculated by subtracting the initial volume of water from the final volume after submerging the iron. \[ \text{Volume of iron} = 63.3 \, \text{mL} - 50.4 \, \text{mL} = 12.9 \, \text{mL} \] Now, we can calculate the density: \[ D = \frac{97.4 \, \text{g}}{12.9 \, \text{mL}} \approx 7.56 \, \text{g/mL} \] Next, we find the percent error using the formula: \[ \% \text{-error} = \left(\frac{\text{actual} - \text{experimental}}{\text{actual}}\right) \times 100\% \] Plugging in the values: \[ \% \text{-error} = \left(\frac{7.874 - 7.56}{7.874}\right) \times 100\% \approx 3.98\% \] So, Stewart's percent error in measuring the density of the iron is approximately **3.98%**.