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mojhsa [17]
1 year ago
5

. an ionic compound is composed of 34.95 g of iron and 15.05 g of oxygen. find the empirical formula of this compound.

Chemistry
1 answer:
Contact [7]1 year ago
3 0

Empirical formula of ionic compound is FeO. In which the composition of atoms is 1 : 1.

Empirical formula of an ionic compound is defined as the formula which gives whole number ratio of atoms of various elements present in molecule of compund.

mass of iron in compound = 34.95 g

molar mass of iron = 55.8 g

mass of oxygen in compound = 15.05 g

molar mass of oxygen = 32 g

number of moles of iron present in the compound are ratio of mass of iron in compound/ molar mass of iron

number of moles of iron in compound= 34.95 / 55.8 = 0.6263 ~ 1

number of moles oxygen in compound= 15.05/ 32 = 0.473 ~ 0.5

the ratio of the number of oxygen atoms to number of iron atoms present in one formula unit of iron compund is 2×0.5 / 1 = 1 : 1

Hence , the required empirical formula of iron compound is FeO.

To learn more about Emiprical formula, refer:

brainly.com/question/1439914

#SPJ4

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1) For the following reaction, 8.44 grams of carbon (graphite) are allowed to react with 9.63 grams of oxygen gas.C(graphite)(s)
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The answers to the question are

(a) 13.24 g

(b) (O₂)

(c) 4.8252 g

(2) 0.662 M/L

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(a) To solve the question we write the equation as follows

C + O₂ → CO₂

That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide

number of moles of graphite = 8.44/12 = 0.703 moles

number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of  carbon to fore 0.3009 moles of  CO₂

The maximum mass of carbon dioxide that  can be formed = mass = moles × molar mass

= 0.3009×44 = 13.24 g

(b) The formula for  the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

(2) The molarity is given by number of moles per liter of solution, therefore

molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

Therefore the moles in 125 mL = (8.3 × 10⁻² M)/(125 mL) = (8.3 × 10⁻² M)/(0.125 L) = 0.662 M/L

3 0
3 years ago
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