(Given, molar mass in $\text{g\ mol}^{-1}$ : $\text{O} = 16, \text{Mg} = 24, \text{P} = 31$)
Phosphorus Percentage Calculation in Compound X
This solution details the calculation for determining the percentage of phosphorus in compound (X) based on the gravimetric analysis yielding magnesium pyrophosphate ($\text{Mg}_2\text{P}_2\text{O}_7$).
1. Molar Mass Calculations
Calculate the molar mass of magnesium pyrophosphate ($\text{Mg}_2\text{P}_2\text{O}_7$) using the given atomic masses:
- Molar mass of Mg = $24 \text{ g mol}^{-1}$
- Molar mass of P = $31 \text{ g mol}^{-1}$
- Molar mass of O = $16 \text{ g mol}^{-1}$
Molar mass of $\text{Mg}_2\text{P}_2\text{O}_7 = (2 \times \text{M}(\text{Mg})) + (2 \times \text{M}(\text{P})) + (7 \times \text{M}(\text{O}))$
Molar mass of $\text{Mg}_2\text{P}_2\text{O}_7 = (2 \times 24) + (2 \times 31) + (7 \times 16)$
Molar mass of $\text{Mg}_2\text{P}_2\text{O}_7 = 48 + 62 + 112 = 222 \text{ g mol}^{-1}$.
2. Mass of Phosphorus Determination
Determine the total mass of phosphorus (P) present in one mole of $\text{Mg}_2\text{P}_2\text{O}_7$.
Mass of P in $\text{Mg}_2\text{P}_2\text{O}_7 = 2 \times \text{M}(\text{P}) = 2 \times 31 = 62 \text{ g mol}^{-1}$.
3. Phosphorus Percentage Calculation
Calculate the actual mass of phosphorus derived from the $1.79 \text{ g}$ of magnesium pyrophosphate obtained:
Mass of P = Mass of $\text{Mg}_2\text{P}_2\text{O}_7 \times \frac{\text{Mass of P}}{\text{Molar mass of Mg}_2\text{P}_2\text{O}_7}$
Mass of P = $1.79 \text{ g} \times \frac{62 \text{ g mol}^{-1}}{222 \text{ g mol}^{-1}}$
Mass of P $\approx 0.50 \text{ g}$.
Finally, calculate the percentage of phosphorus in the original $1.00 \text{ g}$ sample of compound (X):
Percentage of P = $ \frac{\text{Mass of P}}{\text{Mass of Compound X}} \times 100\% $
Percentage of P = $ \frac{0.50 \text{ g}}{1.00 \text{ g}} \times 100\% $
Percentage of P = $50\%$.
Final Answer Conclusion
The calculated percentage of phosphorus in compound (X) is $50\%$. This matches the nearest integer value provided in the options.
Similar Questions
- In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is ______ %.
(Aqueous tension at 300 K is 15 mm). Given below are two statements:
Statement I: The dipole moment of R-CN is greater than R-NC and R-NC can undergo hydrolysis under acidic medium to produce
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Statement II : Compound (Y), shown below, has two carbons with $\text{sp}^3$ hybridization, one carbon with $\text{sp}^2$ and one carbon with $\text{sp}$ hybridization
In the light of the above statements, choose the correct answer from the options given below :
Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :
Important Questions from Organic Chemistry
- In Dumas method for estimation of nitrogen, 0.50 g of an organic compound gave 70 mL of nitrogen collected at 300 K and 715 mm pressure. The percentage of nitrogen in the organic compound is ______ %.
(Aqueous tension at 300 K is 15 mm). Given below are two statements:
Statement I: The dipole moment of R-CN is greater than R-NC and R-NC can undergo hydrolysis under acidic medium to produce
Statement II: R-CN hydrolyses under acidic medium to produce a compound which on treatment with $\text{SOCl}_2$, followed by the addition of $\text{NH}_3$ gives another compound(x). This compound (x) on treatment with $\text{NaOCl/NaOH}$ gives a product, that on treatment with $\text{CHCl}_3\text{/KOH/}\Delta$ produces R-NC
In the light of the above statements, choose the correct answer from the options given below- Given below are two statements:
Statement I: Cross aldol condensation between two different aldehydes will always produce four different products.
Statement II: When semicarbazide reacts with a mixture of benzaldehyde and acetophenone under optimum pH, it forms a condensation product with acetophenone only.
In the light of the above statements, choose the correct answer from the options given below - $0.25 \text{ g}$ of an organic compound "A" containing carbon, hydrogen and oxygen was analysed using the combustion method. There was an increase in mass of $\text{CaCl}_2$ tube and potash tube at the end of the experiment. The amount was found to be $0.15 \text{ g}$ and $0.1837 \text{ g}$, respectively. The percentage of oxygen in compound A is ______ %. (Nearest integer)
(Given: molar mass in $\text{g mol}^{-1}$ $\text{H} : 1, \text{C} : 12, \text{O} : 16$) - Grignard reagent $\text{RMgBr (P)}$ reacts with water and forms a gas $\text{(Q)}$. One gram of Q occupies $1.4 \text{ dm}^3$ at STP. (P) on reaction with dry ice in dry ether followed by $\text{H}_3\text{O}^+$ forms a compound $\text{(Z)}$. 0.1 mole of $\text{(Z)}$ will weigh _________ g. (Nearest integer)
