Phosphorus-32 has a half life of 14.3 days, an atomic mass of 32.0. what mass is left from 175mg after 35 days

Question

what mass of phosphorus-32 is left from an original sample of 175mg of Na3PO4 after 35.0 days?

Answer 1

14.3 days times X = 35. This gives the number of half-lives. Divide 175 by the number of half-lives. The atomic mass is irrelevant.

Answer 2

First, convert mass of Na3PO4 to moles:

MM = 3 * 22.99 + 32.0 + 4 * 16.00 = 164.97 g/mol

moles = 0.175 g / 164.97 g/mol = 0.00106 mol Na3PO4 = 0.00106 mol P

now, for the decay calculations:

find k from the formula k = -ln(1/2) / halflife = 0.693 / 14.3 days = 0.0485 /day

next, use this formula:

ln(Nf) - ln(Ni) = -kt

where Nf = final moles, Ni = initial moles, k = rate constant and t = time

plug n' chug...

ln(Nf) - ln(0.00106) = -0.0485 * 35 = -1.70

ln(Nf) = -1.70 + ln(0.00106) = -8.55

Nf = e^-8.55 = 0.000194 mol

last but not least, convert back to mass

0.000194 mol * 32.0 g/mol = 0.00622 g = 6.22 mg