A mixture of Mg CO3 and MgCO3.3H20 has a mass of 2.883g. After heating to drive

Question

A mixture of Mg CO3 and MgCO3.3H20 has a mass of 2.883g. After heating to drive off all the water the mass is 2.427g. What is the mass percent of MgCO3.3h20 in the mixture? someone on yahoo answers got the correct answer but I don't understand the math behind it. Can someone help me with this: MgCO3 * 3H2O is customarily written with a dot in the middle of the line between the formula for the salt and the number of water molecules of hydration that come with the hydrated form of the salt. The difference in mass was caused entirely by the loss of water, so: (2.883 g - 2.427 g) / (18.01532 g H2O/mol) x (1 mol MgCO3*3H2O / 3 mol H2O) x (138.3597 g MgCO3*3H2O/mol) / (2.883 g) = 0.4049 = 40.49% MgCO3 * 3H2O by mass *update: I've found where my mistake was*

Explanation / Answer

mass of water lost = 2.883 - 2.427 = 0.456 g

No of mol of water lost = 0.456/18 = 0.0253 mol

1 mol MgCO3.3H20 = 3 mol H2O

no of mol of MgCO3.3H20 present in sample = 0.0253/3 = 0.00843 mol

so that,

mass of MgCO3.3H20 present in sample = 0.00843*138.4 = 1.167 g

mass percentage of MgCO3.3H20 = mass of MgCO3.3H20/ total mass *100

                                                       = (1.167/2.883)*100

                                = 40.5%

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