The chemical formula of yellowcake is U3O8. How many uranium atoms are present in 74.3 mg of yellowcake? (Avogadro's number is N0 = 6.022e+23)

Answer 1

The Molar Mass of U3O8 is 842.09 grams, so the starting point here would be taking the given mass and converting over into moles

0.0743 g x 1 mol / 842.08 g = 8.823 * 10^(-5) mol

From there then you want to multiply by Avogadro's number to convert over into atoms

8.823 * 10^(-5) mol * 6.022 x 10^23 molecules/mole = 1.594 x 10^20 atoms

meaning that the number of uranium atoms in 74.3 mg of yellowcake is approximately 1.594 x 10^20 atoms

Answer 2

First, note that the mass is in mg, so use 0.0743g. The molar mass of U is 238.03; multiply this by 3 and add 16*8 for oxygen giving 842.1g molar mass of the compound. Then use dimensional analysis:

(0.0743g U₃O₈) * (1 mol U₃O₈ / 842.1g U₃O₈) * (6.022 x 10²³ particles/ 1 mol whatever) * (3 atoms U / 1 particle of U₃O₈) = 1.59 x 10²⁰ atoms U in 74.3 mg of U₃O₈.