Question

Perform the following calculations; express each answer in exponential form and with the appropriate number of significant figures. (a) \(0.406 \times 0.0023=\) (b) \(0.1357 \times 16.80 \times 0.096=\) (c) \(0.458+0.12-0.037=\) (d) \(32.18+0.055-1.652=\)

Short answer

(a) \(9.3 \times 10^{-4}\), (b) \(2.2 \times 10^{-1}\), (c) \(5.4 \times 10^{-1}\), (d) \(3.058 \times 10^{1}\)

Step-by-step solution

Solve (a) - Perform the Calculation

Perform multiplication for \(0.406 \times 0.0023\), which will give the result 0.000934.

Solve (a) - Determine Number of Significant Figures

The number of significant figures in the final calculation must be the same as the number with the least significance in the original numbers. Here, \(0.0023\) has two significant figures, hence the answer will also be in two significant figures, which results in 0.00093.

Solve (a) - Convert to Exponential Form

Write the answer in exponential form to give \(9.3 \times 10^{-4}\).

Solve (b) - Perform the Calculation

Perform multiplication for \(0.1357 \times 16.80 \times 0.096\), which will give the result 0.22058.

Solve (b) - Determine Number of Significant Figures

Based on the rule for multiplication and division, the number of significant figures in the final calculation must equal the smallest number of significant figures in the original numbers. Hence in this case, the answer will be in two significant figures 0.22.

Solve (b) - Convert to Exponential Form

Convert the decimal to exponential notation, which will yield \(2.2 \times 10^{-1}\).

Solve (c) - Perform the Calculation

Perform the operations in order \(0.458 + 0.12 - 0.037\), which will yield 0.541.

Solve (c) - Determine Number of Significant Figures

The answer for addition and subtraction should be reported to the lowest decimal place contained in the original numbers. Hence, the answer should be reported to the hundredth place to yield 0.54.

Solve (c) - Convert to Exponential Form

Write the calculation in exponential form, which will yield \(5.4 \times 10^{-1}\).

Solve (d) - Perform the Calculation

Perform the operations in order \(32.18+0.055-1.652\), which will yield 30.583.

Solve (d) - Determine Number of Significant Figures

Because we are adding and subtracting, the final answer should be reported to the hundredth decimal place, 30.58.

Solve (d) - Convert to Exponential Form

The final answer will be in exponential form, \(3.058 \times 10^{1}\).

Key concepts

Exponential Notation

Exponential notation is a handy way to express very large or small numbers. It's often used in chemistry due to the vast range of measurements you encounter. By writing numbers in the form of \(a \times 10^n\), you simplify expressions and make them easier to manage.
This format, where \(a\) is a number called the coefficient, and \(n\) is an integer, helps in clearly showing the number of significant figures.
For example, the number 0.00093 is expressed as \(9.3 \times 10^{-4}\). This method ensures that each value is shown in its purest form without any unnecessary digits. Always remember that the exponent \(n\) indicates how many places the decimal point moves to convert the coefficient to the standard form.

• Positive \(n\): Moves decimal to the right.
• Negative \(n\): Moves decimal to the left.

Mastering exponential notation can make calculations less cumbersome and allow for consistent reporting of your final results.

Multiplication and Division in Chemistry

In chemistry, precision in reporting results is crucial, and this is where understanding significant figures comes into play, especially in multiplication and division. The rule is relatively simple: the number of significant figures in the result is determined by the original number with the fewest significant figures.
Consider the example of multiplying \(0.1357 \text{ and } 16.80 \text{ and } 0.096\). The smallest significant figure number here is from 0.096, which has two significant figures. Hence, any product from this operation should also be expressed with two significant figures, resulting in \(2.2 \times 10^{-1}\).
This ensures that precision is not overestimated in your results, maintaining the integrity of the data, especially when measurements have uncertainties inherent to the experiment. Being diligent in this practice is vital for accurate data reporting and interpretation.

Addition and Subtraction in Chemistry

When adding or subtracting numbers in chemistry, attention shifts slightly from significant figures to decimal places. The least precise measurement—the one with the fewest decimal places—determines the precision of the result.
Take for instance the operation \(32.18 + 0.055 - 1.652\). Here, the smallest decimal place is at the hundredth place, dictated by 32.18, which controls how many decimal places we report in our answer.
Therefore, your result should coincide with this precision level, yielding 30.58.

• Always align decimals before performing operations.
• Round the result to the precision of the least precise measure.

Understanding this concept helps ensure that the results you report are as precise as possible, without falsely increasing the apparent accuracy of your data.