EXPERIMENTAL PROBLEM n. 11 GROUP By: Eemeli, Olli, Younes, Giovanni, Federico
You receive a sample of a salt formed by an "s-block" metal and an halogen. The amount of this sample contains as many halogen ions as there are in 1,00 grams of sodium chloride. You have to recognize the metal and the halogen in that salt.
Vi viene assegnato un campione di sale formato da un metallo del "blocco s" e da un alogeno. La quantità di questo campione contene tanti ioni alogenuro quanti ve ne sono in 1,00 g di cloruro di sodio. Dovete riconoscere il metallo e l'alogeno nel sale.
Saat suolanäytteen, joka koostuu alkali- tai maa-alkaliryhmän metallista (=s-ryhmä) ja halogeenista. Näyte sisältää yhtä monta halogeeni-ionia kuin 1,00 grammassa natriumkloridia (NACl) on kloridi-ioneita. Tehtävänä on tunnistaa, mistä metalli- ja halogeeni-ioneista suola koostuu.
1. Useful knowledge (reference data, concepts and principles)
1.1 Colours of alkali metals to the bunsen flame:
| Lithium |
Sodium |
Potassium |
Calcium |
Strontium |
Barium |
| deep purple |
yellow_orange |
lilac |
orange+red sparkle |
red sparkle |
green-yellow |
1.2 Atomic relative mass of sodium and chlorine: Na 22.99; Cl 35.45
1.3 Formula Weight of NaCl: 58.44
1.4 Formula of a salt formed by Me+ and X- ions: MeX
1.5 Formula of a salt formed by Me2+ and X- ions: MeX2
2. Main idea
we recognize the metal with the flame test and the halogen through a calculation based on the weight of the salt
3. Planning
1. we take the bowl that contained the salts assigned
2.kindle the flame and take the wires of nickel
3. we clean the threads with acid
4. take salt with wire
5. and put salt on the flame
6.conclude that salt is
4. Report
| Italian 29-03-2010 repeated 19-04-2010 |
Finnish 31.03.2010 |
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tub full + salt = 1.88 g
empty tub = 2.88 g
salt flame colour = yellow/orange/reddish
Flame test |
empty box = 13,5g
box with salt = 14,8g
weight of salt = 1,3g
salt colour = yellow/orange
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5. Elaboration and Interpretation
| Italian |
Finnish |
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Data elaboration-calculation from raw data
deductions and interpretations from calculated data
conclusions: answering problem request. Other outcomes.
The metal can be Na or Ca
2,88-1,88=1,00 g
if the metal is sodium our salt would be NaCl;
this is impossible because nobody was assigned this salt.
If the metal is Ca we have CaX2. So we calculate:
1/58,44=0,0171 g number of X- ions
number of Ca2+ ions = 0,0171/2= 0.00855
mass of Ca=40·0,00855=0,342g
mass of X=1,00g-0,342=0,658g
atomic mass of X = 0,658/0,0171= 38.5
we deduce that the halogen is Chlorine
because its expected atomic mass would be 35,5,
that is very near to our outcome.
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weight of salt = 1,3 g
Sodium or Calsium.
1g : 58,44 = 0,0171
0,0171:2 = 0,00855 atoms of halogen X
1,3 - 0,00855 = 1,29145
1,29145 : 0,0171 = 75,52339181
halogen = Br = Bromi
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Common conclusions
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Questions & Answers
1. Why are atoms emitting light at the flame?
Answer 1 change color because the atoms are first excited and then they relax themselves reemitting an impulse of light
2. How did you recognize the metal element in the salt?
2. Come avete riconosciuto l'elemento metallico nel sale?
Answer 2
Risposta 2
3. What measure permitted you to deduce which halogen was in the salt?
3. Quale misura vi ha permesso di dedurre quale alogeno è presente nel vostro sale?
Answer 3
Risposta 3
4. Does your salt contain more cations or anions?
4. Il vostro sale contiene più cationi o anioni?
Answer 4
Risposta 4
5. Does your salt contain more mass as halide ions or as metal ions or the same?
5. Il vostro sale contiene una maggiore massa di ioni alogenuro, di ioni del metallo o la stessa massa?
Answer 5
Risposta 5
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