Question

ATP is an important molecule in living cells. sample with a mass of 0.8138 g was analyzed and found to contain 0.1927 g C, 0.02590 g H, 0.1124 g N, and 0.1491 g P. The remainder was O. Determine the empirical formula of ATP. Its formula mass was found to be 507 g mol-1. Determine its molecular formula

Answer 1

Mass of carbon = 0.1927 g

Mass of hydrogen = 0.02590 g

Mass of nitrogen = 0.1124 g

Mass of phosphorus = 0.1491 g

Mass of Oxygen = 0.8138 - ( 0.1927 + 0.02590 + 0.1124 + 0.1491 ) = 0.3337 g

Moles of C = $\frac{0.1927}{12.011} = 0.0160 mol$

Moles of H = $\frac{0.02590}{1.008} = 0.0257$ mol

Moles of O = $\frac{0.3337}{15.999} = 0.0209 mol$

Moles of P = $\frac{0.1491}{30.974} = 0.00481 mol$

Moles of N = $\frac{0.1124}{14.01} = 0.00802 mol$

So, the empirical formula will be: $C_{0.0160} H_{0.0257} O_{0.0209} N_{0.00802} P_{0.00481}$

To make it a whole divide the moles of all elements by 0.00481.

Therefore, $C_3_3 H_5_5 O_4_3 N_1_7P_1$.

Molecular mass of ATP = 507 g/mol

Empirical formula = 164.063 g/mol

Molecular formula:

$\frac{507}{164} = 3$

So, molecular formula is $(C_3_3 H_5_5 O_4_3 N_1_7P_1)_3$ = $C_1_0 H_1_6 O_1_3 N_5P_3$

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