In an organic compound, phosphorus is estimated as M*g_{2}*P_{2}*O_{7} (Molar mass= 222 g). If 0.24 g of an organic compound gave 0.44 g of M*g_{2}*P_{2}*O_{7} What is the Dercentage of Phosphorus in the compound ? Atomic mass: Mg = 24u P = 31u and O = 16 u)
Question
In an organic compound, phosphorus is estimated as M*g_{2}*P_{2}O_{7} (Molar mass= 222 g). If 0.24 g of an organic compound gave 0.44 g of Mg_{2}*P_{2}*O_{7} What is the Dercentage of Phosphorus in the compound ? Atomic mass: Mg = 24u P = 31u and O = 16 u)
Solution
To find the percentage of phosphorus in the compound, we need to follow these steps:
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First, we need to find the molar mass of Mg2P2O7. Using the atomic masses given, we can calculate this as follows: (224) + (231) + (7*16) = 222 g/mol.
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Next, we need to find the mass of phosphorus in Mg2P2O7. Since there are two atoms of phosphorus in each molecule, this is 2*31 = 62 g/mol.
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Now, we can find the mass of phosphorus in the 0.44 g of Mg2P2O7. We do this by multiplying the mass of Mg2P2O7 by the ratio of the mass of phosphorus to the molar mass of Mg2P2O7. This gives us (0.44 g * 62 g/mol) / 222 g/mol = 0.123 g of phosphorus.
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Finally, we can find the percentage of phosphorus in the original compound. We do this by dividing the mass of phosphorus by the mass of the original compound and multiplying by 100. This gives us (0.123 g / 0.24 g) * 100 = 51.25%.
So, the percentage of phosphorus in the compound is 51.25%.
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