Question

Find the weight of P in 1.5254g and 1.5936g of MgNH₄PO₄

Answer 1

Atomic masses of Mg, N, H, P, O can be given as follows:

Atomic mass of Mg = 24.3050

Atomic mass of N = 14.0067

Atomic mass of H = 1.00794

Atomic mass of P = 30.973762

Atomic mass of O = 15.9994

Thus, mass of MgNH4PO4 can be calculated as follows:

Mass of MgNH4PO4 = 1 x (mass of Mg) + 1 x (mass of N) + 4 x (mass of H) + 1 x (mass of P) + 4 x (mass of O)

Mass of MgNH4PO4 = 1 x 24.3050 + 1 x 14.0067 + 4 x 1.00794 + 1 x 30.973762 + 4 x 15.9994

Mass of MgNH4PO4 = 137.314822 g/mol

Therefore, 137.314822 g of MgNH4PO4 contains 30.973762 g of P. Weight of P in 1.5254g and 1.5936 of MgNH4PO4 can be calculated as follows:

Weight of P in 1.5254g of MgNH4PO4 = [(30.973762 g of P)/( 137.314822 g of MgNH4PO4)] x 1.5254g of MgNH4PO4

Weight of P in 1.5254g of MgNH4PO4 = 0.34408g of P

Weight of P in 1.5936g of MgNH4PO4 = [(30.973762 g of P)/( 137.314822 g of MgNH4PO4)] x 1.5936g of MgNH4PO4

Weight of P in 1.5254g of MgNH4PO4 = 0.35946g of P

Thus, 1.5254g of MgNH4PO4 contains 0.34408g of P and 1.5936 of MgNH4PO4 contains 0.35946g of P