The anticancer drug Platinol (cisplatin), Pt (NH3) 2Cl2, reacts with the cancer cell's DNA and interferes with its growth. (a) What is the mass % of platinum (Pt) in Platinol? (b) If Pt costs 851/9, how many grams of Platinol can be made for $1.00 million (assume that the cost of Pt determines the cost of the drug) ?

(a) = 65.018% (b) = 1.807,83 grams

Explanation:

(a)

The mass % of platinum (pt) in platinol can be obtained applying the general percentage formula (portion of interest / total), in this case the platinol its composed of a single platinum (Pt) mole with molecular weight of 195.088 g/mol, two nitrogen (N) mole of 28.012 g/mole each. six hydrogen (H) mole of 1.007 g/m and two chlorine (Cl) of 35.453 g/mole

N = 14.006 g/mole x 2 = 28.012

H = 1.0078 g/mole x 6 = 6.046

Cl = 35.453 g/mole x 2 = 70.906

Pt = 195. 088 g/mole x 1 = 195.088

the molecular weight of platinol its = 300.052g

% of Pt in platinol its = ((195.088 g/mole) / (300.052 g/mole)) * 100 = 65.018%

(b)

Assumption: 851/9 its in fact 851 $/gram

to know how many grams of Platinum can be bought it's necessary to divided the total money between the price of platinum, knowing how many grams can be bought and knowing the percentage of platinum in platinol, it's possible to know how many grams of platinol can be made from $1.00 million

grams of platinum $1'000.000 / (851$/g) = 1.175,09 grams of Pt

grams of platinol =

1.175,09 grams of pt / 0.65018 (grams of pt/grams of platinol)

grams of platinol = 1.807,83 grams of platinol