Problem 22 - General Chemistry: Principles and Modern Applications

Question

Perform the following calculations; express each number and the answer in exponential form and with the appropriate number of significant figures. (a) \(\frac{320 \times 24.9}{0.080}=\) (b) \(\frac{432.7 \times 6.5 \times 0.002300}{62 \times 0.103}=\) (c) \(\frac{32.44+4.9-0.304}{82.94}=\) (d) \(\frac{8.002+0.3040}{13.4-0.066+1.02}=\)

Step-by-Step Solution

Answer

The calculated results are (a) \(9.8 \times 10^4\), (b) \(2.48 \times 10^1\), (c) \(0.4461\), and (d) \(0.578\).

Step 1: Convert to Exponential Form and Calculate (a)

Convert \(320\), \(24.9\), and \(0.080\) to exponential form as \(3.20 \times 10^2\), \(2.49 \times 10^1\), and \(8.0 \times 10^{-2}\) respectively. Now perform the operation \(\frac{3.20 \times 10^2 \times 2.49 \times 10^1}{8.0 \times 10^{-2}}\) to get the answer which equals \(9.8 \times 10^4\).

Step 2: Convert to Exponential Form and Calculate (b)

Convert \(432.7\), \(6.5\), \(0.002300\), \(62\), and \(0.103\) to exponential form as \(4.327 \times 10^2\), \(6.5 \times 10^0\), \(2.3 \times 10^{-3}\), \(6.2 \times 10^1\) and \(1.03 \times 10^{-1}\) respectively. Now perform the operation \(\frac{4.327 \times 10^2 \times 6.5 \times 10^0 \times 2.3 \times 10^{-3}}{6.2 \times 10^1 \times 1.03 \times 10^{-1}}\), which equals \(2.48 \times 10^1\) considering significant figures.

Step 3: Add and Subtract before Division (c)

Add the values \(32.44\) and \(4.9\), and subtract \(0.304\) to get \(37.036\). Then divide this value by \(82.94\) to get \(0.4461\), considering number of significant figures.

Step 4: Follow the sequence of operations (d)

Add \(8.002\) and \(0.3040\) to get \(8.306\), and subtract \(0.066\) from \(13.4\), then add \(1.02\) to the result, resulting \(14.36\). Now perform the operation \(\frac{8.306}{14.36}\) to get \(0.578\), considering the significant figures.

Key Concepts

Exponential Form

Exponential form is a way to express numbers as a product of a number (also known as the coefficient) and a power of ten. This method is particularly useful in chemistry when dealing with both very large and very small numbers, as it simplifies calculations and helps maintain accuracy.

• For example, the number 320 can be expressed in exponential form as \(3.20 \times 10^2\). This means 3.20 is multiplied by the power of 10 squared, or 100, which equals 320.
• Similarly, the number 0.080 becomes \(8.0 \times 10^{-2}\). The negative exponent indicates a fraction, thus 0.080 is equivalent to \(\frac{8.0}{100}\).

Expressing numbers in exponential form clarifies the number of significant figures and allows for easier manipulation, especially in multiplication and division operations. It is important to always maintain the correct number of significant figures in the coefficient to ensure accurate scientific reporting.

Calculation

Calculating with numbers in chemistry requires precision and attention to significant figures. When performing calculations involving several steps, the number of significant figures in the final result should match the measurement with the fewest significant figures.

• In multiplication or division, the rule is to round the final answer to the same number of significant figures as the factor with the fewest significant figures. For instance, while solving \(\frac{320 \times 24.9}{0.080}\), we consider the lowest number of significant figures among 320 (2 significant figures), 24.9 (3 significant figures), and 0.080 (2 significant figures), rounding the result to 2 significant figures.
• In arithmetic operations (addition and subtraction), rounding is different. The result is rounded off to the least precise decimal place from the numbers involved in the calculation. In the division example \(\frac{32.44+4.9-0.304}{82.94}\), we add and subtract first, using the least precise measurement (4.9) after the decimal to round the result correctly.

Proper calculation maintains scientific accuracy and ensures reliable experimental outcomes.

Division in Chemistry

Division operations in chemistry often involve numbers expressed in scientific notation, requiring careful manipulation of both coefficients and exponents.

• When dividing numbers in exponential form, divide the coefficients as regular numbers and subtract the exponents. For example, dividing \(\frac{3.20 \times 10^2 \times 2.49 \times 10^1}{8.0 \times 10^{-2}}\) involves dividing 3.20 by 8.0 and subtracting exponents: \(2 + 1 - (-2)\). The correct handling of powers of 10 simplifies calculations and yields the correct exponential form result.
• Similarly, when handling multiple division steps, ensure to simplify each part sequentially while maintaining significant figures. This method is crucial when dealing with compound fractions in chemistry calculations, often leading to balanced and precise solutions.

Understanding the division process in chemistry ensures numerical accuracy, which is vital when deducing chemical concentrations, reaction yields, or any formula derivations.