3a Lab Report

3.Introduction- In order to do this experiment you must know certain terms and formulas. Molar mass is defined as one mole of this substance. The molar mass is what we are trying to find.One equation use d to find molar mass is MW=m/n. Another thing needed to know is the Ideal Gas Law. The equation used for this is PV=nRT.

Using elemental analysis to determine the percent mass composition of each element in a compound is the first step in creating an empirical formula. There are many different types of elemental analysis, but in this experiment gravitational analysis and Beer’s Law are used. Elemental analysis is first used to find the moles of each element, then converted to mass, and then the percent mass of the element in the product is found (2).

Weight the mass for the different percent of zinc and copper mixtures, which have 11 groups.

Molar Mass is basically in grams per mole for example salt has a molar mass of 5.8g per mole. The mass of an element refers to 6.0221415x1023 molecules of the substance. Molar Mass helps us figure out how many grams per mole we need to do this project and it helps us know how much we need to put into the solution. If you don’t Find

With the mass of oxygen found, 0.02 grams, the percent compositions of magnesium and oxygen can be found by dividing their masses by the total mass of the new compound:

Because of the Law of Definite Proportions, when the elements of magnesium and oxygen chemically react to form a compound they do so in definite proportions by weight. The Law of Definite Proportions states that if elements form a compound, they must combine in definite proportions by weight. This means that all atoms form chemical compounds by combining in whole number ratios. Take for example the elements of hydrogen and oxygen: they can combine to make water (H2O) but in other conditions can combine to make hydrogen peroxide (H2O2). In both examples the ratio stays the same. Likewise, in this experiment the elements magnesium and oxygen combine in whole-number ratios. We have hypothesized that magnesium and oxygen

The relationship between mass of the product and mass of the reactant taken directly from the balanced chemical equation and molar masses of the involved species is used in this experiment to find the mass of the unknown sample. In this experiment the unknown sample is found out ,which will be on of the following salts:NaHCO3,Na2CO3,KHCO3K2CO3.Only the fixed mass of reactant react together to produce fixed mass of products.

To prepare a quantitative solution, you need to know the weight of the substance and the quantity of the solution. For example, you have 40 grams of NaOH (Sodium Oxide) in 1000mL of water. The amount of water and weight of the substance makes a Mole. one mole is equal to 1000mL of water and 40 grams of NaOH and varies by the amount of water you have but the weight of the substance must also change. To make a correct solution, you need to know the atomic mass of the substance and how much water you have in mL or L. If there are multiple elements, you need to add the combined weight of all elements (EX. NaOH= 23+16+1+40 grams.) and then divide the weight by the mass. To make a solution, you should use a beaker or flask that can measure at least

A chemical equation demonstrates the quantity of atoms of every component that make up the compound. The atomic proportion of components in the compound is the same as the molar proportion of those components. This information can be applied to solve for the mass of one mole of a compound by including the molar masses of the constituent components.

In the method of continuous variations the total number of moles of reactants is kept constant for the series of measurements. Each measurement is made with a different mole ratio of reactants. A mole ratio is ratio between the amounts in moles of any two compounds involved in a chemical reaction. Mole ratios are used as conversion factors between products and reactants in many chemistry problems.

Although the tin did not show any signs of reaction some of the results show there was still a 0.01 grams loss in mass. More accurate scales and more experiments conducted to give a better understanding to see if there was any actual loss in the tin. The iron on the other did show signs of reactions and had a 0.01 grams mass loss. To get more accurate results of how much mass was actually lost there would need to be more accurate scales. Signs of reaction also occurred with the zinc. There was an average of 0.033 grams mass loss. If more accurate scales are unavailable an alternative to the experiment would be to increase the concentration of acid or increase the time period of the reaction (say three days).

The purpose of this lab was to become familiar with the three different balances and two different methods used to find the weight and mass of chemicals and compounds in the ChemLab program. The lab was performed by using three different types of balances, and the direct weighing and weighing by difference methods.

This experiment was performed to become familiar with chemical formulas and the processes for the chemical reactions. By breaking down the chemicals, we were able to obtain certain reactants that were already part of the product. In three trials, the zinc chloride’s mass ranged from 1.16 g to 1.62 g. Mass of the chlorine found in the zinc chloride ranged from 0.66 g to 0.96 g. In the three trials of the copper sulfide lab, the copper sulfide’s mass ranged from 2.38 g to 3.18 g. The mass of the sulfur in this lab ranged from 0.64 g to 1.51g.

This is how you can calculate the average atomic mass; multiply the fraction by the mass number for each isotope, then add them together.

In this lab, an empirical formula for the Magnesium oxide was investigated . The empirical formula of the first and second trials was calculated, and it was MgO. We used the average mass the MgO, then subtracted it from the mass of the crucible to find the mass of MgO in g. After we got the mass of MgO, we subtracted it from the m of mg to get the mass of O. We calculated the mass present of the O, then we found the molar mass of it. The molar mass of O was calculated by multiplying its mass by grams to 1 mol dividing by its standard atomic weight in (g). We found the molar mass of the Mg, then we divided the smallest value by each element in order to get a small whole number. The whole numbers for each element that are Mg and O lead us to