I've been on this problem for days and really need help. Work would be amazing, but even the theory on how to do this would be helpful. Thanks!

Earth's surface area is 5.10x10^8 km^2; it's crust has a mean thickness of 35 km and a mean density of 2.8 g/cm^3. The two most abundant elements in the crust are oxygen (4.55x10^5 g/t, where t stands for metric ton; 1 t= 1000 kg) and silicon (2.72x10^5 g/t), and the two rarest nonradioactive elements are ruthenium and rhodium, each with an abundance of 1x10^-4 g/t. What is the total mass of each of these elements in Earth's crust?

## Answers (2)

The crust will have a total mass of 5.10 x 10^8 km^2 * 35 km (thickness).

This is 1.8 x 10^10 km^3. (Two significant figures)

How many cubic cm in a cubic km? Well, it takes 100,000 cm to make a km. This is 10^5. Cubing, the conversion from km^3 to cm^3 adds 10^15, or 1.8 x 10^25 cm^3.

Now, how much does that weigh? 1.8 x 10^25 * 2.8 = 5.0 x 10^25 g or 5.0 x 10^19 t.

Multiply each of your numbers by the total weight (in t) to get the total weight of the element.

For O, I get 2.2 x 10^25 g (or about 44%)

For Si, I get 1.4 x 10^25 g (or about 28%)

For Ru and Rh, I get 5.0 x 10^15 g (or about 0.0001%)

Check my math, especially for the last calculation, because I'm awful with a calculator.

5.1 X 10^8 km^2 X 35 km = 1.785X10^10 km^3

1.785 X 10^10 km^3 X (1000 m/km)^3 X (100 cm/m)^3 = 1.785 X 10^25 cm^3

11.785 X 10^25 cm^3 X 2.8 g/cm^3 = 5.0 X 10^25 g

5.0 X 10^25 g X 1 kg/1000 g X 1 t / 1000 kg = 5.0 X 10^19 t

5.0 X 10^19 t X (2.72 X 10^5 g/t) = 1.4 X 10^25 g (which you could convert back to metric tons, if you like)

5.0 X 10^19 t X (1X10^-4 g/t) = 5X10^15 g