Abstract: In this lab our main objective is to find the empirical formula of MgO, magnesium oxide. To do this first we have to make sure when we burn Mg in the crucible and it reacts to with O. This lab experiment Mg is complicated by another factor. Mg is so reactive that it reacts with N in the air so some of the Mg which is suppose to react with O will react will N instead. This obstacle can be overcome by simply adding water to the Mg3N2 to yield MgOH2 and NH3. This will undo the reaction so O can react with Mg to make MgO and then we can get the empirical formula of MgO.
Introduction: The correct formula for magnesium oxide is MgO which is a 1.0 to 1.0 ratio. Meaning for every 1 mol of magnesium, Mg, there one mol of O, oxygen. However, …show more content…
Would extra mass in the crucible cause the "mass of oxygen" to come out too high or too low? Explain. A: Too high, because the mass of magnesium won’t change. Anything that result in the mass of oxygen will increase the mass.
2. The correct formula for magnesium oxide is MgO, a 1.0 to 1.0 ratio. But sometimes in this experiment the ratio of Mg to O comes out too low. (Example: 0.9 Mg to 1.0 O) In that case, it means that there was too much oxygen relative to the mass of magnesium. At other times it comes out that the ratio is too large. An example would be: 1.2 to 1.0 (Mg to O). In such a case it must be that there has been too little oxygen (or too little weight at the end of the experiment, which registers as too little oxygen.) In each case below, decide whether the situation described would lead to a calculated ratio of too much oxygen, or too little oxygen, and explain your reasoning. (The calculated ratio contains the mistake.) a. Putting in more water than is needed for reaction 3, and then not drying out this excess water. A: Too little oxygen will present in the crucible because oxygen can not dissolve easily in …show more content…
Once you have calculated the standard deviation of the student results for the class, explain its significance by stating the 68% and 95% confidence intervals. (We have 95% confidence that the true value of the mole ratio lies between and , assuming only random errors.)
Conclusion: We did not complete this lab and the end result was off. This happened due to factors like over heating the crucible and breaking it. It put us really behind and in the process of changing the crucible some mass of MgO was lost. However, we did get the reaction we were expecting but not the right result. In conclusion, our empirical formula for MgO was not calculated because we did not finish the experiment. We did not reach to the point where we had to extract all the Mg3N2 by adding distilled water and reheating the Mg so the rest of O2 can react with it and give us pure
Introduction: The correct formula for magnesium oxide is MgO which is a 1.0 to 1.0 ratio. Meaning for every 1 mol of magnesium, Mg, there one mol of O, oxygen. However, …show more content…
Would extra mass in the crucible cause the "mass of oxygen" to come out too high or too low? Explain. A: Too high, because the mass of magnesium won’t change. Anything that result in the mass of oxygen will increase the mass.
2. The correct formula for magnesium oxide is MgO, a 1.0 to 1.0 ratio. But sometimes in this experiment the ratio of Mg to O comes out too low. (Example: 0.9 Mg to 1.0 O) In that case, it means that there was too much oxygen relative to the mass of magnesium. At other times it comes out that the ratio is too large. An example would be: 1.2 to 1.0 (Mg to O). In such a case it must be that there has been too little oxygen (or too little weight at the end of the experiment, which registers as too little oxygen.) In each case below, decide whether the situation described would lead to a calculated ratio of too much oxygen, or too little oxygen, and explain your reasoning. (The calculated ratio contains the mistake.) a. Putting in more water than is needed for reaction 3, and then not drying out this excess water. A: Too little oxygen will present in the crucible because oxygen can not dissolve easily in …show more content…
Once you have calculated the standard deviation of the student results for the class, explain its significance by stating the 68% and 95% confidence intervals. (We have 95% confidence that the true value of the mole ratio lies between and , assuming only random errors.)
Conclusion: We did not complete this lab and the end result was off. This happened due to factors like over heating the crucible and breaking it. It put us really behind and in the process of changing the crucible some mass of MgO was lost. However, we did get the reaction we were expecting but not the right result. In conclusion, our empirical formula for MgO was not calculated because we did not finish the experiment. We did not reach to the point where we had to extract all the Mg3N2 by adding distilled water and reheating the Mg so the rest of O2 can react with it and give us pure