### Introduction

** **The ideal gas law is used to define how gasses typically act. It is not precise, as gasses do not usually act ideally, but it works for most laboratory conditions. The ideal gas law is defined by PV = nRT (pressure in atm * volume in L = moles * constant * temperature in K). There is also a rule that states the total pressure of a system is equal to the sum of its partial pressures. Using this information, a small piece of metal magnesium is reacted with hydrochloric acid. The reaction is represented by the equation Mg (s) + 2HCl (aq) –> MgCl_{2 }(aq) + H_{2} (g). Using the sum of partial pressures rule and the ideal gas law to find moles, the molar mass of Mg (s) can be calculated only knowing the pressure of H_{2 }(g).

### Experimental

First, a strip of magnesium metal was obtained and weighed on an analytical balance. This weight was recorded to three significant figures. Next, the magnesium strip was wrapped around the tip of a copper wire, and was then encaged by the same copper wire by wrapping the wire around the magnesium. Following this, the tip of a buret was calibrated and this volume was recorded. Next, about 3 mL of concentrated HCl was put in the empty buret with the stopcock closed, and distilled water was then added until the buret was almost full. The copper wire with the magnesium was then lowered into the buret and held in place with a one-holed rubber stopper to make sure the wire would stay in place and would not fall to the bottom. Distilled water was squirted into the one-holed rubber stopped to make sure the buret was completely filled. The buret was then quickly inverted and clamped into a 600 mL beaker half filled with tap water. The water level on the buret was recorded after production of gas ceased. Finally, the distance between the height of water in the beaker and water in the buret was also recorded. The process was repeated for one more trial.

### Results

Data | Trial 1 | Trial 2 |

Mass of Mg sample, g | 0.0406 | 0.0384 |

Volume of uncalibrated portion of buret, mL | 1.20 | 1.20 |

Final buret reading, mL | 9.40 | 11.52 |

Volume of hydrogen, mL | 41.80 | 39.68 |

Temperature of hydrogen, ºC | 23 | 22 |

Barometric pressure, mm Hg | 762.76 | 762.76 |

Differences in water levels between buret and beaker, mm H_{2}O | 135 | 144.5 |

Pressure difference of water levels, mm Hg | 10.0 | 10.7 |

Aqueous vapor pressure at temperature of hydrogen, mm Hg | 21.2 | 19.8 |

Pressure of hydrogen after correction for difference in water levels and vapor pressure, mm Hg | 731.6 | 732.3 |

Pressure converted to atm, atm | 0.9626 | 0.9636 |

Absolute temperature, K | 296 | 295 |

Volume of hydrogen, L | 0.04180 | 0.03968 |

Moles of hydrogen gas produced, moles | 0.001656 | 0.001579 |

Calculated molar mass of Mg, g/mole | 24.52 | 24.32 |

Actual molar mass of Mg, g/mole | 24.31 | 24.31 |

Percent error | 0.8638 | 0.04113 |

Average percent error: 0.4525

### Calculations

** **In order to find the volume of hydrogen, I took the total volume of the buret (50 mL plus the calibrated part) and subtracted my final reading. Using the results from my first trial, 50.0 mL + 1.20 mL – 9.40 mL = 41.80 mL. To find the pressure difference of water levels, I took the distance in difference of water levels, multiplied by the density of water, and divided by the density of Hg. From the first trial, 135 mm (1.0 g/mL) / 13.5 g/mL = 10.0 mm. The aqueous vapor pressure was found using a chart. At 23 ºC, the aqueous vapor pressure is 21.2 torr (or 21.1 mm Hg). Using the equation P_{atm} = P_{H2O (l)} + P_{H20 (g)} + P_{H2 (g)}, I could solve for the pressure of H_{2} (g). Subbing in the numbers from the first trial, 762.76 torr = 21.2 torr + 10.0 torr + P_{H2 (g)}. P_{H2 (g)} = 731.6 torr. To find atmospheres, I multiplied by 1 atm/760 torr. 731.6 torr (1 atm/760 torr) = 0.9626 atm. I then rearranged the ideal gas law to find moles of hydrogen produced. n = PV/RT, n = 0.9626 atm * 0.04180 L / (0.0821 L atm mole^{-1} K^{-1} * 296 K), n = 0.001656 moles. In the equation for the reaction, there are an equal number of moles of hydrogen and magnesium used. To find the molar mass of magnesium, I took the mass of the magnesium used divided by the number of moles used. From trial 1, 0.0406 g / 0.001656 moles = 24.52 g/mol. To find percent error, I took the absolute value of the true value minus my calculated value divided by the true value and multiplied by 100%. Using trial 1, ( | 24.31 – 24.52 | ) / 24.31 * 100% = 0.8638%. Finally, to find the average percent error, I added the two percents and divided by two: (0.8638% + 0.04113%) / 2 = 0.4525%.

### Discussion/Conclusions

My results are very good. An average percent error of 0.4525% is extremely small, so I feel that I performed the experiment well and had some good luck. I felt that when performing the experiment, I could have gotten some error from my temperature readings. The temperature seemed to fluctuate a couple degrees during the experiment and I thought that would negatively affect my final results. I also thought I could have gotten error from my difference in water levels measurement because it was hard to hold the ruler steady and read the ruler at the same time. Lastly, since the ideal gas law is not precise, error could have come from that. Other than those questionable factors, the experiment went smoothly.