   Chapter 4, Problem 88GQ

Chapter
Section
Textbook Problem

Boron forms a series of compounds with hydrogen, all with the general formula BxHy. B x H y ( s )   +  excess O 2 ( g ) → x 2 B 2 O 3 ( s ) + Y 2   H 2 O(g) If 0.148 g of one of these compounds gives 0.422 g of B204 when burned in excess 02, what is its empirical formula?

Interpretation Introduction

Interpretation:

The empirical formula of boron compound in the given reaction has to be determined.

Concept introduction:

• Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
• Equation for finding Molecular formula from the empirical formula,

MolarmassEmpiricalformula mass × Empirical formula

Numberofmole=GivenmassofthesubstanceMolarmass

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
Explanation

0.148g of boron compound BxHy burns in oxygen then give 0.422gofB2O3.

Balanced equation for the given reaction is,

BxHy(s)+excessO2(g)x2B2O3(s)+y2H2O(g)

0.148g of boron compound BxHy burns in oxygen then give 0.422gofB2O3.

Mass of boron (in gram) in 0.422g of B2O3 is,

0.422gofB2O369.619g/molB2O3×1molB1molB2O3×22.6122g/molB=0.131g

Mass of oxygen (in gram) in 0.148g of BxHy is,

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