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Alchen [17]
3 years ago
11

The compound cisplatin, pt(nh3)2cl2 , has been studied extensively as an antitumor agent.

Chemistry
1 answer:
Nataliya [291]3 years ago
5 0

a. Elemental percent composition is the mass percent of each element in the compound.

The formula for mass elemental percent composition = \frac{mass of element}{mass of compound}   (1)

The molecular formula of cisplatin is Pt(NH_3)_2Cl_2.

The atomic weight of the elements in cisplatin is:

Platinum, Pt = 195.084 u

Nitrogen, N = 14.0067 u

Hydrogen, H = 1 u

Chlorine, Cl = 35.453 u

The molar mass of Pt(NH_3)_2Cl_2 = 195.084+ (2\times 14.0067)+(6\times 1)+(2\times 35.453) = 300.00 g/mol

The mass of each element calculated using formula (1):

- Platinum, Pt %

\frac{195.084}{300.00} \times 100 = 65.23%.

- Nitrogen, N%

\frac{2\times 14.0067}{300.00} \times 100 = 9.34%

- Hydrogen, H%

\frac{6\times 1}{300.00} \times 100 = 2.0%

- Chlorine, Cl%

\frac{2\times 35.453}{300.00} \times 100 = 23.63%

b. The given reaction of cisplatin is:

K_2PtCl_4(aq)+2NH_3(aq)\rightarrow Pt(NH_3)_2Cl_2(s)+2KCl(aq)

According to the balanced reaction, 1 mole of K_2PtCl_4 gives 1 mole of Pt(NH_3)_2Cl_2.

Now, calculating the number of moles of K_2PtCl_4 in 100.0 g.

Number of moles = \frac{given mass}{Molar mass}

Molar mass of K_2PtCl_4 = 2\times 39.0983+195.084+4\times 35.453 = 415.093 g/mol

Number of moles of K_2PtCl_4 = \frac{100 g}{415.093 g/mol} = 0.241 mole.

Since, 1 mole of K_2PtCl_4 gives 1 mole of Pt(NH_3)_2Cl_2. Therefore, mass of cisplatin is:

0.241 mole\times 300 g/mol = 72.3 g

For mass of KCl:

Molar mass of KCl = 39.0983 + 35.453 = 74.55 g/mol

Since, 1 mole of K_2PtCl_4 gives 2 mole of KCl. Therefore, mass of KCl is:

0.241 mole\times 74.55 g/mol\times 2 = 35.93 g


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