a. 5.40g of bromine reacted with 8.58g of iodine. What is anempiracal formula of

Question

a. 5.40g of bromine reacted with 8.58g of iodine. What is anempiracal formula of the compound that was formed? b. What is the molecular formula of this compound if we knowthat its molar mass is 206.8g? c. What is the percent composition of this compound? a. 5.40g of bromine reacted with 8.58g of iodine. What is anempiracal formula of the compound that was formed? b. What is the molecular formula of this compound if we knowthat its molar mass is 206.8g? c. What is the percent composition of this compound?

Explanation / Answer

Molecular Mass (mm) Br=79.909g/mol  5.4g*1mol/79.909g=.06758mol Br mm I =126.904g/mol    8.54g*1mol/126.904g=.06761mol I These are both row 7 elements (they need one bond to have afull shell) so it makes sense that the stoichiometry is 1:1 so, BrI was formed. The molecular formula of mm 206.8g results from the 1:1combination of Br and I if you add the molar mass of Br with the molar mass of I youget 126.904+79.904=206.81g, you have exactly one mole of Br combining with exactly 1 mol of I. so the molecular formula is 1mol Br+1mol I------>1molBrI The percent composition is 50% Br and 50% I the percent by mass is 38.6%Br and 61.4% I (79.909/206.81*100%=38.6% and 129.9/206.81*100%=61.4%)

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