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Molar Mass of Butane

1-HgO(s) Hg(l) + O2(g)
Consider the unbalanced equation above. A sample of impure mercury(II) oxide is heated and the HgO decomposed completely. If 680. mL of O2 is collected by displacement of water at a barometric pressure of 680.0 mm Hg and 25.0°C, what mass of HgO was originally present? The vapor pressure of water is 23.8 mm Hg at 25.0°C. Use molar masses with at least as many significant figures as the data given.

2-Na(s) + H2O(l) H2(g) + NaOH(aq)
Consider the unbalanced equation above. What volume of H2 is produced at 860. mm Hg and 24.0°C when 19.2 g of sodium reacts with excess water? Use molar masses with at least as many significant figures as the data given.

3HgO(s) Hg(l) + O2(g)
Consider the unbalanced equation above. A sample of impure mercury(II) oxide is heated and the HgO decomposed completely. If 595 mL of O2 is collected by displacement of water at a barometric pressure of 720.0 mm Hg and 25.0°C, what mass of HgO was originally present? The vapor pressure of water is 23.8 mm Hg at 25.0°C. Use molar masses with at least as many significant figures as the data given.

4-Na(s) + H2O(l) H2(g) + NaOH(aq)
Consider the unbalanced equation above. What volume of H2 is produced at 600. mm Hg and 32.4°C when 16.6 g of sodium reacts with excess water? Use molar masses with at least as many significant figures as the data given.

Data Table

Measurement = Trial 1 Trial 2
Initial mass of lighter (g) = 13.5 and 13.5
Final mass of lighter (g) = 13.7 and 13.7
Volume of butane (ml) = 48 and 54
Barometric Pressure (inches of Hg) = 30.26 and 30.23
Temperature of water (°C)= 19 and 19

Part two Calculated Value for Trial 1 and Trial 2: above is what I recorded use it to answer the follwoing:

1-Mass of butane used (g)
2-Volume of butane used (L)
3-Temperature of water (in Kelvin)
4-Use 273.15 for the conversion.
5-Barometric Pressure (mm Hg)
6-Vapor Pressure of water (mm Hg)
Click here to look up vapor pressure of water
7-Pressure of butane (mm Hg)(Ptotal = Pbutane + Pwater vapor)
8-Pressure of butane (atm)
9-Moles of butane used (PV=nRT) Use 0.08206L.atm/mol.K for R.
10- Molar mass of butane (g/mol)(Keep 3 sigfigs.)
11-Average Molar Mass(g/mol)(Keep 3 sigfigs.)
12-Theoretical molar mass of butane, C4H10 (g/mol)(Keep 2 decimal places.)
13-Percent error(Keep 2 sigfigs.).

Solution Preview

See the attached file.

1- HgO(s) Hg(l) + O2(g)
Consider the unbalanced equation above. A sample of impure mercury(II) oxide is heated and the HgO decomposed completely. If 680. mL of O2 is collected by displacement of water at a barometric pressure of 680.0 mm Hg and 25.0°C, what mass of HgO was originally present? The vapor pressure of water is 23.8 mm Hg at 25.0°C. Use molar masses with at least as many significant figures as the data given.

2HgO →2Hg+O2

P(O2)=680.0-23.8=656.2 mmHg
PV=nRT
n(O2)=PV/(RT)=656.2/760*0.680/(0.0821*(273.15+25.0))=0.024 mol
Based on the chemical reaction, the molar amount of HgO consumed: 2*n(O2)=0.048 mol
Mass: 0.048*216.6=10.40 g.
Therefore, there are 10.40 g of HgO present at the beginning.

2-Na(s) + H2O(l) H2(g) + NaOH(aq)
Consider the unbalanced equation above. What volume of H2 is produced at 860. mm Hg and 24.0°C when 19.2 g of sodium reacts with excess water? Use molar masses with at least as many significant figures as the data ...

Solution Summary

The solution discusses the molar mass of butane.

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