Hydrochlorimetry Lab Report

Improved Essays
By adding Hydrochloric acid into solid Magnesium, in Data table #2, when Hydrochloric acid was first been added to solid Magnesium, the ribbon immediately start to bubble and producing fumes at the same time with a burning sound. Magnesium start to turn into a white liquid form. After the bubble start to disappear, more Hydrochloric acid was added, and the reactions repeats. Until none of the shiny silver Magnesium ribbon can be seen, and there is no more bubble, the reaction is over. After placing the evaporating dish on a hot plate on a scale of 7, by the data from Data table #2, the white liquid that contains Magnesium, Chlorine and Hydrogen start to release fumes. As time passed, the liquid start bubble, and the liquid start to become …show more content…
To find the moles of Magnesium, dimensional was used to multiply the molar mass of Magnesium to its measured mass, the final result is 5.3475936 x 10-3 moles.
0.13g Mg×(1mol/(24.31g))=5.3475936×〖10〗^(-3)
After adding Hydrochloric acid and been heated on the hot plate, the total mass of evaporating dish plus Magnesium Chloride is 44.96 grams. By subtracting the mass of Magnesium ribbon 0.13 grams from the total mass 44.96 grams, the mass of Magnesium Chloride was found to be 0.56 grams.
44.96-44.40=0.56g
After knowing the mas of Magnesium chloride, by subtracting the mass of Magnesium ribbon from it, the mass of Chloride was found to be 0.43
…show more content…
0.43g Cl×(1mol/(35.45g))=1.121297602×〖10〗^(-2) After knowing the moles of Magnesium and Chlorine, their mole ratio can be calculated by dividing the smaller moles, and then multiply a positive whole number to get the final empirical formula ratio of 9 Mg with 4 Cl.
(0.0121297602)/(0.0053475936)=2.26 (0.0053475936)/(0.0053475936)=1 Cl=2.26×4=9.07 Mg=1×4=4
As we know each mass of Magnesium and Chloride, the percent composition of Magnesium can be calculated when divide the mass of magnesium from the mass of Magnesium Chloride.
(0.13)/(0.56)=23.21428571%
After knowing the percent composition of Magnesium, and knowing the theoretical value as 25.5%, the percent error can be calculated by applying the formula.
(25.5-23.2142857)/(25.5)×100=8.963585451%

Related Documents

  • Improved Essays

    This lab makes use of the reaction excess powdered calcium carbonate and different concentrations limiting hydrochloric acid in order to determine the effect of changing concentration on the rate of the reaction. Students will carry out 3 trials of 5 experiments each trial. The five different experiments are for the various concentrations of hydrochloric acid (0.2, 0.4, 0.6, 0.8, 1.0 mol/L). First measure out approximately 2 grams of powdered calcium carbonate using a weighing boat and analytical balance. Then, measure out 30 mL of 0.2 M hydrochloric acid into a volumetric flask.…

    • 198 Words
    • 1 Pages
    Improved Essays
  • Improved Essays

    Light the Bunsen Burner and then use the tongs to put the magnesium ribbon into the flame. Do not look at the ignited flame because the intensity of the flame could burn the retina in the eyes. Then you pull the magnesium ribbon out of the flame. 2 Mg(s) + O2(g) = 2MgO(s)…

    • 793 Words
    • 4 Pages
    Improved Essays
  • Improved Essays

    The purpose of this lab was to determine the formula of copper chloride hydrate by dehydration through heating, rehydration through adding water back in, the reaction of aluminum with the copper chloride hydrate to produce solid copper, and the weight of this copper. To accomplish this, a petri dish was weighed and 1.01 g of copper chloride hydrate was added to the dish. The solid salt crystals were broken up to make sure that all of the hydrate would be heated the same. Next, the petri dish and hydrate were heated upon a hot plate for roughly 15 minutes, until all the salt crystals were brown instead of blue. This indicated that all of the water had been burned out of the hydrate, leaving dehydrated copper chloride.…

    • 494 Words
    • 2 Pages
    Improved Essays
  • Improved Essays

    Clean Evaporating Dish

    • 247 Words
    • 1 Pages

    As per lab manual, we weigh an evaporating dish, then place an unknown substance on it, and weigh it as well. From this, we calculate the sample’s weight. We then set the dish up over a burner and heat the sample until the white fumes it produces are no longer expelled. When the dish cools, we weigh it again and calculate the amount of NH4Cl that was present in the mixture, due the amount of weight lost. Following this, we add 25 mL of water to the solid present in the evaporating dish and mix it.…

    • 247 Words
    • 1 Pages
    Improved Essays
  • Decent Essays

    If heating magnesium in the presence of oxygen and they react completely, then the ionic compound’s empirical formula can be found using this equation by calculating the masses of magnesium and oxygen in this equation: Mg + O2 → MgxOy Since an ionic compound is the hopeful result of this lab, an ionic compound is a chemical compound comprising ions, which ions are an atom or molecule with a net electric charge due to the loss or gain of an electron. Ions can be negatively charged and called anions, or positively charged and called anions. But in this lab, many concepts and techniques will be used to get this compound. The main technique that will be used is heating to a constant mass.…

    • 309 Words
    • 2 Pages
    Decent Essays
  • Improved Essays

    The empirical formula for the ratio of moles H2O over moles anhydrous salt is MgSO4 - 7 H2O. The percent composition of the hydrate of water is 50.8%. When heating hydrates to a very high temperature, the physical bonds are broken and the water is then evaporated from the substance. Prior to heat being added to the magnesium sulfate heptahydrate, the substance was very brittle in tiny little pieces. However, after heat was added to the substance for several minutes, the substance became very powdery and more condensed than before.…

    • 220 Words
    • 1 Pages
    Improved Essays
  • Improved Essays

    Purpose: The purpose of this lab is to prove and discern the empirical formula of copper oxide using the Law of Conservation of Mass. This states that in measuring the mass of each component of copper oxide and determining the percent composition of each the copper and the oxygen within copper oxide it is possible to find the ratio of the elements within the compound. Procedure: In order to conduct this lab, 25.1mL of distilled water then 25.5mL of 3M HCl must be added 2.11g of copper oxide.…

    • 477 Words
    • 2 Pages
    Improved Essays
  • Improved Essays

    Lab Report On 100-Memol

    • 263 Words
    • 2 Pages

    A strip of magnesium (Mg) ribbon was cut with a scissor. Said piece had a length of 6 cm, and a weight of 0.071 g. The magnesium piece was rolled into a compact ribbon. After that, with copper turnings the mg piece was covered. Next, 700-mL of water was placed into a 1000-mL beaker. Then, the copper-wrapped magnesium was placed in the beaker.…

    • 263 Words
    • 2 Pages
    Improved Essays
  • Improved Essays

    Atomic mass: the mass of an atom of a chemical element expressed in atomic mass units; approximately equivalent to the number of protons and neutrons in the atom (the mass number) or to the average number allowing for the relative abundances of different isotopes Molar mass: a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance Mole: the amount of a substance contained in 6.02 x 10²³ particles of that substance Percent composition: the percent by mass of each element present in a compound Empirical formula: a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms Molecular formula: a formula giving the number of atoms…

    • 418 Words
    • 2 Pages
    Improved Essays
  • Great Essays

    With the progression of each trial, bubbles were generated on the magnesium strip as a result of the reaction between the reactants. This not only caused the magnesium strip to float to the top of the acid but also formed a block between the magnesium and hydrochloric acid. These occurrences very possibly caused a slower than expected reaction between the metal and acid. 2. It is highly probable that, from the time the magnesium strip was sanded down to the time it was dropped into the HCl, the magnesium strip could have formed a thin layer of oxidized magnesium.…

    • 1908 Words
    • 8 Pages
    Great Essays
  • Improved Essays

    The Element Of Magnesium

    • 672 Words
    • 3 Pages

    The element Magnesium got its name when it was discovered. It is a silvery white metal that is highly reactive. Its chemical symbol is Mg and atomic number is 12. Its atomic mass is relatively 24. Therefore, it has 12 protons, 12 neutrons, and 12 electrons.…

    • 672 Words
    • 3 Pages
    Improved Essays
  • Improved Essays

    The fire will be accelerated by the oxygen in the air making it more intense and dangerous. The more the surfaces of the magnesium are exposed the higher the intensity of the fire. It is advisable not to use water in extinguishing magnesium fires however small they are. The equation for the chemical process involved in the re-ignition of the fire is Mg + 2H2O > Mg(OH)2 + H2 and also MgO + H2O >…

    • 956 Words
    • 4 Pages
    Improved Essays
  • Improved Essays

    Magnesium Oxide Lab Report

    • 1193 Words
    • 5 Pages

    The burning of magnesium in open air causes a chemical reaction that forms the compound, magnesium oxide. This reaction occurs due to the heat that provides energy to activate the reaction. A blinding white flame will burn as the reaction occurs. In this lab, magnesium is burned in an open crucible over a Bunsen burner several times until the magnesium ceases to burn. This process allows for the magnesium to completely react with the oxygen found in the open air, forming the chemical compound magnesium oxide.…

    • 1193 Words
    • 5 Pages
    Improved Essays
  • Superior Essays

    Double Salt Lab Report

    • 1556 Words
    • 7 Pages

    Experimental: Materials – double salt (self-made & EIU Chemistry Stockroom), 6 M HCl (EIU Chemistry Stockroom), 10% BaCl2, 6 M HNO3 (EIU Chemistry Stockroom) Equipment – Ocean Optics Spectrometer USB4000, Logger Pro Procedure – In order to find the mass percent of the sulfate, the sulfate needed to separate itself from the rest of the double salt. To do this, experimenters began by weighing out 0.998 g of the double salt into a 250-mL beaker. They then used the graduated cylinder to add about 150 mL of deionized water and 2 mL of the 6 M HCl into the 250-mL beaker. As they added the deionized water to the salt, it dissolved relatively quickly with the help of a glass stirring rod.…

    • 1556 Words
    • 7 Pages
    Superior Essays
  • Decent Essays

    Molecular weights, also known as molar mass is the mass of a molecule in grams per mole, or the sum of the component atomic masses. A volatile liquid is a substance easily evaporated at normal temperatures. The molecular weight can be found by determining the mass of a vapor under known conditions of temperature, volume and pressure. Once the pressure, volume and temperature are determined the moles of the gas can be calculated by using the Ideal Gas Law, in which PV =nRT. In this experiment, the molar mass of an unknown volatile liquid will be identified using the ideal gas…

    • 102 Words
    • 1 Pages
    Decent Essays