From Empirical Formulas

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Presentation transcript:

From Empirical Formulas Molecular Formulas From Empirical Formulas

A compound containing only carbon, hydrogen, and oxygen is 63 A compound containing only carbon, hydrogen, and oxygen is 63.16% C and 8.77% H. It has a molar mass of 114 g/mol. What is its empirical and molecular formula? Write the formula in terms of grams Convert g to moles by dividing by the molar mass of each element. Convert the small numbers by dividing each mole quantity by the smallest mole quantity. All quantities are whole numbers. Divide the known molar mass by the mass of one mole of the empirical formula. The result produces the integer by which you multiply the empirical formula to obtain the molecular formula.

From Combustion Analysis Empirical Formulas From Combustion Analysis

X =#moles of C or #moles of CO2 A 3.489 g sample of a compound containing C, H, and O yields 7.832 g of CO2, and 1.922 g of water upon combustion. What is the empirical formula of the compound? CxHyOz + O2  CO2 + H2O X =#moles of C or #moles of CO2 Y = #moles of H which is twice #moles of water. Z = # moles of O. Obtain g of O and then convert to moles. Obtain empirical formula.

Formulas of Hydrates

CuSO4•XH2O(s)  CuSO4(s) + XH2O(g) When 21.91 g of a hydrate of copper(II) sulfate is heated to drive off the water, 14.00 g of anhydrous copper(II) sulfate remain. What is the formula of the hydrate? CuSO4•XH2O(s)  CuSO4(s) + XH2O(g) Calculate the g of water by subtracting the grams of copper(II) sulfate from g of the hydrate. Find moles of H2O/moles of CuSO4 will give you X.