Question

The anticancer drug Platinol (cisplatin), \(\operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\), reacts with a cancer cell's DNA and interferes with its growth. (a) What is the mass \% of platinum (Pt) in Platinol? (b) If Pt costs \(\$ 51 / \mathrm{g},\) how many grams of Platinol can be made for \(\$ 1.00\) million (assume that the cost of Pt determines the cost of the drug)?

Short answer

a) 65.01% b) 30159.08 g

Step-by-step solution

Calculate the molar mass of Platinol

First, find the molar mass of each element in the chemical formula \(\text{Pt(NH}_3\text{)}_2 \text{Cl}_2\). The atomic masses are: \(\text{Pt} = 195.08 \text{ g/mol}, \text{N} = 14.01 \text{ g/mol}, \text{H} = 1.01 \text{ g/mol}, \text{Cl} = 35.45 \text{ g/mol}\). Calculate as follows: \(\text{Molar mass} = \text{Pt} + 2(\text{N} + 3\text{H}) + 2\text{Cl}\) \(\text{Molar mass} = 195.08 + 2(14.01 + 3 \times 1.01) + 2 \times 35.45 = 195.08 + 2(17.04) + 70.90 = 195.08 + 34.08 + 70.90 = 300.06 \text{ g/mol}\)

Calculate the mass % of platinum in Platinol

The mass percent of platinum \( \text{Pt} \) is calculated by dividing the mass of platinum by the molar mass of Platinol and multiplying by 100. \(\text{Mass \%} = \frac{195.08}{300.06} \times 100 \%\ = 65.01 \%\)

Determine grams of Platinol for $1 million

With platinum costing \( \$51/\text{g} \), calculate how many grams of platinum can be bought with \( \$1,000,000 \) by dividing the total money by the cost per gram: \( \frac{1,000,000}{51} = 19607.84 \text{ g of Pt} \). Given that platinum comprises 65.01 \% of Platinol's mass, the total mass of Platinol can be calculated by dividing the grams of platinum by the mass percent (converted to decimal): \( \frac{19607.84}{0.6501} \approx 30159.08 \text{ g of Platinol} \).

Key concepts

mass percent

Mass percent is a way of expressing the concentration of an element in a compound. It's a ratio of the mass of a specific element to the total mass of the compound, multiplied by 100 to get a percentage.

For example, to find the mass percent of platinum (Pt) in Platinol, we first calculate the molar mass of Platinol. The molar mass is the sum of the atomic masses of all the atoms in the formula. Platinol's formula is \(\text{Pt(NH}_3\text{)}_2 \text{Cl}_2\):

\ \text{Molar mass} = 195.08 \text{ (atomic mass of Pt)} + 2 \times (14.01 \text{ (atomic mass of N)} + 3 \times 1.01 \text{ (atomic mass of H)}) + 2 \times 35.45 \text{ (atomic mass of Cl)} = 300.06 \text{ g/mol} \ \text{Mass percent of Pt} = \frac{195.08 \text{ (mass of Pt)}}{300.06 \text{ (molar mass of Platinol)}} \times 100 \ = 65.01 \%.

This calculation tells us that platinum makes up 65.01% of the mass of Platinol.

Remember: Understanding mass percent helps compare the amount of an element in different compounds and is essential for purity analysis and formulation in chemistry.

chemistry calculations

Chemistry calculations involve using mathematical methods to solve problems related to chemical reactions, compounds, and formulas. These calculations often include finding molar masses, mass percents, and stoichiometric ratios.

Let's revisit the Platinol problem. To find out how many grams of Platinol can be made for \(\text{\textdollar 1.00 million}\) given the cost of \(\text{Pt}\) determines the cost of the drug, follow these steps:

• Determine grams of Pt that can be bought with \(\text{\textdollar 1.00 million}\).
• Calculate the total mass of Platinol using the mass percent of Pt.

First, find the mass of Pt that can be bought:
\ \frac{1,000,000 \text{\textdollar}}{{51 \text{\textdollar/g}}} = 19607.84 \text{ g of Pt}

Next, since Pt comprises 65.01% of Platinol's mass:
\ \frac{19607.84 \text{ g of Pt}}{0.6501} \ ≈ 30159.08 \text{ g of Platinol}

This means you can make approximately 30159.08 grams of Platinol with \(\text{\textdollar 1.00 million}\) worth of platinum.

• Always double-check your units and make sure to convert percentages to decimals when needed.
• Use precise atomic masses for accurate results.

Regular practice with these calculations makes understanding chemistry much easier!

stoichiometry

Stoichiometry is a branch of chemistry focused on the quantitative relationships between reactants and products in a chemical reaction. It often involves using balanced equations and molar ratios to make predictions.

Here's how to apply stoichiometry in the context of the Platinol problem: when a reaction involves \(\text{Pt(NH}_3\text{)}_2 \text{Cl}_2\), stoichiometry helps us understand the proportions in which the elements combine and react.

For instance:

• The formula indicates that each molecule of Platinol contains 1 atom of Pt, 2 molecules of \(\text{NH}_3\), and 2 atoms of Cl.
• Balancing these components ensures that we can accurately calculate the amount of each element in the compound and the overall molar mass.
• Stoichiometry also aids in converting mass of reactants to mass of products using molar ratios.

In steps:

• Begin by finding molar masses based on atomic masses of each atom within the molecule.
• Use these molar masses to perform conversions between mass, moles, and molecules.
• Rely on mole ratios derived from the balanced chemical equation for accurate stoichiometric calculations.

Practice and familiarity with stoichiometric principles help you solve complex problems related to chemical reactions and predict outcomes effectively!