Question

dimensional analysis worksheet - densities
name______________________
densities are commonly communicated as how many grams (g) a cubic centimeter (cm³) of the material would weigh. for liquids and gasses ml is often used instead of cm³.
1.000 ml of aluminum would have a mass of 2.699 g or expressed as 1.000 ml al = 2.699 g al
because we have an equation we can create 2 conversion factors based on the physical properties of aluminum.
\\(\frac{2.699\\ \text{g}}{1.000\\ \text{ml}} = 1\\) \\(\frac{1.000\\ \text{ml}}{2.699\\ \text{g}} = 1\\)
use the table to create conversion factors to solve the problems below. under each column of your dimensional analysis write the number of significant figures and round your final answer to the appropriate number of significant figures.
example: what is the volume of 5 g of aluminum?
example calculation with dimensional analysis setup
table:
aluminum | 1.000 cm³ | 2.699 g
arsenic | 1.00 cm³ | 5.73 g
calcium | 1.00 cm³ | 1.55 g
carbon | 1.00 cm³ | 2.22 g
chlorine | 1.00 cm³ | 1.51 g
gold | 1.000 cm³ | 19.32 g
iodine | 1.00 ml | 4.93 g
iron | 1.00 cm³ | 7.87 g
lead | 1.000 cm³ | 11.34 g
lithium | 1.0 ml | 0.53 g

1. what is the volume of 5.0 g of arsenic

blank dimensional analysis setup: two vertical lines with a horizontal line, =

1. what is the volume of 20 g of gold

blank dimensional analysis setup: two vertical lines with a horizontal line, =
what is the volume of 13 g of iodine
blank dimensional analysis setup: two vertical lines with a horizontal line, =

1. what is the mass of 10.0 cm³ of arsenic

blank dimensional analysis setup: two vertical lines with a horizontal line, =

1. what is the mass of 200.00 cm³ of gold

blank dimensional analysis setup: two vertical lines with a horizontal line, =

1. what is the mass 1000 ml of iodine

blank dimensional analysis setup: two vertical lines with a horizontal line, =

Explanation:

Problem 1: Volume of 5.0 g of arsenic

Step1: Identify conversion factor

From the table, for arsenic, \(1.00\ \text{cm}^3 = 5.73\ \text{g}\), so the conversion factor is \(\frac{1.00\ \text{cm}^3}{5.73\ \text{g}}\).

Step2: Set up dimensional analysis

We have \(5.0\ \text{g}\) of arsenic. Multiply by the conversion factor:
\(5.0\ \text{g} \times \frac{1.00\ \text{cm}^3}{5.73\ \text{g}}\)

Step3: Calculate and round

\(5.0 \div 5.73 \approx 0.8726\), round to 2 significant figures (since 5.0 has 2 s.f. and 5.73 has 3 s.f.) gives \(0.87\ \text{cm}^3\) (or \(0.87\ \text{mL}\) as \(1\ \text{cm}^3 = 1\ \text{mL}\) for solids here).

Step1: Identify conversion factor

For gold, \(1.000\ \text{cm}^3 = 19.32\ \text{g}\), so conversion factor is \(\frac{1.000\ \text{cm}^3}{19.32\ \text{g}}\).

Step2: Set up dimensional analysis

\(20\ \text{g} \times \frac{1.000\ \text{cm}^3}{19.32\ \text{g}}\)

Step3: Calculate and round

\(20 \div 19.32 \approx 1.035\), round to 2 significant figures (20 has 1 or 2? If 20 is two sig figs, then \(1.0\ \text{cm}^3\); if 1 sig fig, \(1\ \text{cm}^3\). Assuming 2 sig figs: \(1.0\ \text{cm}^3\)).

Step1: Identify conversion factor

For iodine, \(1.00\ \text{mL} = 4.93\ \text{g}\), so conversion factor is \(\frac{1.00\ \text{mL}}{4.93\ \text{g}}\).

Step2: Set up dimensional analysis

\(13\ \text{g} \times \frac{1.00\ \text{mL}}{4.93\ \text{g}}\)

Step3: Calculate and round

\(13 \div 4.93 \approx 2.637\), round to 2 significant figures (13 has 2 s.f.) gives \(2.6\ \text{mL}\) (or \(2.6\ \text{cm}^3\)).

Answer:

\(0.87\ \text{cm}^3\) (or \(0.87\ \text{mL}\))

Problem 2: Volume of 20 g of gold