# Let's Review: The Tools of Quantitative Chemistry - Study Questions - Page 43a: 21

A) A: $2.4\ g/cm^3$ B: $2.703\ g/cm^3$ B) A: $11.18\%$ B: $0.04\%$ C) A: $0.216\ g/cm^3$ B: $0.0016\ g/cm^3$ D) Method B is more precise and more accurate.

#### Work Step by Step

A) Results that are very different from the others (outliers, like 5.811 in method B) should not be included in the calculations. Averages: A: $(2.2+2.3+2.7+2.4)/4=2.4\ g/cm^3$ $\quad\quad\quad\quad$B: $(2.703+2.701+2.705)/3=2.703\ g/cm^3$ B) A: $E=|2.4-2.702|/2.702\times100\%=11.18\%$ $\quad$B: $E=|2.703-2.702|/2.702\times 100\%=0.04\%$ C) A: $\sigma=\sqrt{\frac{(-0.2)^2+(-0.1)^2+(0.3)^2+(0)^2}3}=0.216\ g/cm^3$ $\quad$B: $\sigma=\sqrt{\frac{(0)^2+(-0.001)^2+(0.002)^2}2}=0.0016\ g/cm^3$ D) Method B is more precise, because it had a smaller standard deviation, and it is more accurate because it had a smaller experimental error.

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