A student performed an experiment similar to the one in this module to determine

Question

A student performed an experiment similar to the one in this module to determine the empirical formula of a compound containing chromium (Cr) and O. After heating 0.4550 g of pure Cr wire in an excess of air, the sample was cooled and weighed. The mass of the compound was 0.8749 g. The molar mass of Cr and O are 52.00 g mol -1 and 16.00 g mol -1, respectively.

1. Determine the mass of O in the compound.

2. Determine the number of moles of Cr in the compound.

3. Determine the number of moles of O in the compound.

4. Identify the element that is present in the smaller molar amount in the compound

5. Find the ratio of the number of moles of Cr to the number of moles of the element identified in (4)

6. Find the ratio of the number of moles of O to the number of moles of the element identified in (4)

7. Write the empirical formula for the compound formed.

Explanation / Answer

Lets consider all Cr has fully reacted.

1) 0.8749g - 0.4550g = 0.4199g
Why? - coz in the compound only contains Cr and O2. If the mass of Cr in known so, the rest must be the mass of O2

2) 0.4550g divide by 52.00g/mole = 9.0x10^-3 mole
Why? - You know how to do this.

3) 0.4199g divide by 16.00g/mole = 0.03 mole
Why? - Same as above

4) Check out the value of mole for each of compound (b and c) which one is the smaller
Why? - The value of mole can act as quantity measurement for component

5) 1: 1

f6) 1:3
Why? - Try divide both mole and it should be 0.3..

7) empirical formula = CrO3

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.