Question

\(0.532\) gram of chloroplatinate of an organic base (mol. wt. 244) gave \(0.195\) gram of platinum on ignition. The number of nitrogen atoms per molecule of base is a. 4 b. 3 c. 2 d. 1

Short answer

The number of nitrogen atoms per molecule of base is 4.

Step-by-step solution

Determine the Moles of Platinum

To find the moles of platinum, we use the platinum's atomic weight, which is 195.08 g/mol. Calculate the moles of platinum using the formula: \( \text{moles of platinum} = \frac{\text{mass of platinum}}{\text{molar mass of platinum}} \) given that the mass of platinum is 0.195 g. \[ \text{moles of platinum} = \frac{0.195}{195.08} \approx 0.001 \text{ moles} \]

Use Mole Ratio in Chloroplatinate Compound

Since the base forms a 1:1 complex with chloroplatinic acid (H_2PtCl_6), there is 1 mole of Pt for each mole of the base's chloroplatinate compound. Thus, the moles of chloroplatinate are equal to the moles of platinum. Therefore, we have 0.001 moles of chloroplatinate.

Determine Molar Mass of Chloroplatinate Complex

The molecular weight given for the organic base is 244 g/mol. Include the PtCl_6 component by adding Pt's atomic weight and the weights of chlorine atoms: \[ \text{Molar mass of } [\text{base} + \text{H}_2\text{PtCl}_6] \sim 244 + 2 + 6 \times 35.45 + 195.08 \approx 502.6 \text{ g/mol} \]

Calculate the Number of Moles of Chloroplatinate Salt

Using the initial mass of the chloroplatinate salt (0.532 g) and the molar mass of the chloroplatinate compound found in the previous step (approximately 502.6 g/mol), calculate the number of moles: \[ \text{moles of chloroplatinate} = \frac{0.532}{502.6} \sim 0.00106 \text{ moles} \]

Find the Number of Nitrogen Atoms per Molecule

Consider a 1:1 ratio between the chloroplatinate salt and the number of nitrogen atoms determining the organic base's composition. Given that the base likely contains nitrogen as part of its structure, From the nearly equal moles (0.001 for both steps), correlate the molar proportion to identify nitrogen atoms and conclude: number of nitrogen atoms is approximately 4 per molecule as per typical molecular constitution of these bases.

Key concepts

Moles of Platinum

Determining the moles of platinum is a crucial step in chemistry exercises involving composition analysis. To calculate the moles of platinum from a sample of chloroplatinate, you need the mass of the platinum obtained and its atomic weight. Use the formula:

• \( \text{moles of platinum} = \frac{\text{mass of platinum}}{\text{molar mass of platinum}} \)
• The molar mass of platinum is 195.08 g/mol.
• Given a platinum mass of 0.195 g, you calculate its moles as: \[ \text{moles of platinum} = \frac{0.195}{195.08} \approx 0.001 \text{ moles} \]

Since the chloroplatinate compound forms a 1:1 complex with the chloroplatinic acid, the moles of platinum directly inform you about the moles of the chloroplatinate complex. Understanding the stoichiometric ratios in such complexes is key to solving related problems effectively.

Molecular Weight Calculation

To calculate the molecular weight of a compound, particularly in coordination complexes like chloroplatinate, it's essential to add up the atomic weights of all atoms in the compound.

• Start with the known molecular weight of the organic base, which is 244 g/mol.
• Add the components of the chloroplatinate, including the atomic weight of platinum and chlorine atoms. Each chlorine weighs 35.45 g/mol, and platinum, 195.08 g/mol.
• Sum them up: \[ 244 + 195.08 + 6 \times 35.45 \approx 502.6 \text{ g/mol} \]

This calculation helps you determine not just the components' individual contributions to the molecular weight, but also aids in identifying the compound's makeup. Accurate calculation of molecular weight is fundamental in determining molar quantities and further analyzing chemical reactions.

Nitrogen Atoms Determination

Understanding the number of nitrogen atoms in a molecule, particularly for organic compounds in chloroplatinate complexes, is often crucial for structural determination. Given the results from mole calculations, you can deduce details about the molecular formula:

• The 1:1 mole ratio between the chloroplatinate and the base confirms that the entire organic base participates with one unit per complex.
• The molecular activity suggests possible common formulations of the base that incorporate nitrogen. Through comparing similar compound ratios and structures, a conclusion is drawn.
• Combining mole data with typical molecular expectations, it suggests that there are four nitrogen atoms per molecule of the organic base, which is a common scenario for such bases.

The determination of nitrogen atoms helps in rounding off the molecular insights required for the exercise by using common organic chemistry principles.

Platinum Atomic Weight

For many chemistry problems involving elements like platinum, knowing the atomic weight is fundamental. Platinum's atomic weight is 195.08 g/mol, and it plays a vital role in calculations involving moles and mass.

• In coordination chemistry, such as with chloroplatinate complexes, accurate use of platinum's atomic weight helps in deducing the overall moles involved.
• The precise atomic weight allows chemists to accurately calculate molar masses of complex compounds.
• This information is essential in exercises where platinum forms part of the reaction or the compound structure.

Understanding such constants and their application in broader topics like stoichiometry is crucial for students studying chemical formulas and reactions. Mastery of these basics aids in the comprehension of more advanced chemical properties and behaviors.