Stoichiometry Lab Report

Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029

This calculation being15 grams

What is the molecular formula if the molecular weight is 58.0 g/mole? A compound containing Sulfur and Nitrogen is 69.6% Sulfur by mass. If its molar mass is 184 g/mole: What is the empirical formula What is the

Percentage Composition of Hydrates The Introduction: Hydrates are ionic compounds and are often a type of salt with a definite amount of water as part of their structure. Hydrates are decomposed into anhydrous salt, a no-water compound, and water vapor when heat is applied; water vapor is released from the hydrated compound, which leaves with an anhydrous salt that weighs less than the hydrate. But how do we find the percent of the water that was lost after applying heat to the hydrates and how accurate is finding the percent composition of water using experimentation and a gram weight scale? The purpose of the experiment is to find the percent of water lost by finding the difference between the mass of hydrate and the mass of anhydrous

Which of the following unit is used to indicate mass? a. Cm3 b. Um c. Mg d. mL 21. Which of the following demonstrate a chemical reaction of water?

Eventually using the NaOH and the acid’s consumed moles, the equivalent mass will be determined. Procedure: Part 2: Obtain 45mL of NaOH, and then weigh 0.3-0.4g of the unknown acid (KH2PO4). Dissolve the acid into 20.00mL water.

Lastly, add the molar masses of each element in the compound. 3. Explain

Mole ratio is considered a conversion factor since it helps to convert units with the use of moles. It is used in stoichiometry and other calculations and comes out of the balanced chemical equations. Stoichiometry is the measurement of elements that concerns the chemical quantities produced or taken in a reaction. The process of relating the mass and mole quantities of reactants or products in a reaction. It uses a balanced chemical equation, mole ratio, and sometimes needs mole mass.

75.33 grams Weight of the unknown = 0.23 grams Calculation : 75.33-75.10/0.36x100 = 63.8 % recovery Melting point of

Stoichiometry is a method used in chemistry that involves using relationships between reactants and products in a chemical reaction, to determine a desired quantitative data. The purpose of the lab was to devise a method to determine the percent composition of NaHCO3 in an unknown mixture of compounds NaHCO3 and Na2CO. Heating the mixture of these two compounds will cause a decomposition reaction. Solid NaHCO3 chemically decomposes into gaseous carbon dioxide and water, via the following reaction: 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g). The decomposition reaction was performed in a crucible and heated with a Bunsen burner.

Ideally, every mole of each reagent would be used up, and theoretical yield, we are assuming that every last mole of the reactants would

Empirical Formula of Magnesium Oxide - Lab Report Background Information/Introduction: The aim of this lab is to determine the empirical formula of magnesium oxide by converting magnesium to magnesium oxide. As an alkali earth metal, magnesium reacts violently when heated with oxygen to produce magnesium oxide and magnesium nitride as a byproduct. In order to obtain only magnesium oxide, distilled water was added so that magnesium nitride will react and convert to magnesium hydroxide. Further heating then oxidizes all of the magnesium into magnesium oxide.

Procedure A. Preparation of NaOH solution The molarity of a solution is the ratio of the number of solutes dissolved in a liter of solution. To figure out the needed mass (in grams) of NaOH pellets to be dissolved in a 0.25 L of water, remember that a mole is equivalent to the quotient of mass over the molar mass of the substance. This was used to rearrange the base formula and to derive the mathematical equation of mass in terms of molarity. mass (g) =

℃^(-1)×6.40℃±3.1 %=1337.6 J±4.06 % ∆H=(-1337.6 J±4.06 %) /(0.025 mol ±0.16 %)= -53504 J m〖ol〗^(-1)±4.22 % ∆H=-53504 J m〖ol〗^(-1)±4.22 %÷1000=-54 kJ m〖ol〗^(-1)±4.22 % Conclusion and

The mixture were stirred by using a glass rod until the mixture is fully dissolved. The solution were tested by using calibrated pH meter to get the pH value of the solution. Results and Discussions pH ratio between acid and base: 7.3 = 6.82 + x x = 0.48 0.48 = log ([base])/([acid]) 100.48 =base/acid salt/acid = 3.02 There, 1 acid : 3 base calculate number of mole of acid and base to find the mass : molar = mol/L 50 mM = (mol )/(0.5 L) mol = 25 mol number of mole of NaH2PO4 25/4 = acid = 6.25 mol number of mole of Na2HPO4 25/4 × 3 = salt = 18.75 mol to calculate the mass of the acid and base : Mass of NaH2PO4 (6.25 mol)/(119.98 g/mol)=0.052g