EXPERIMENTAL PROBLEM n. 11 GROUP 10
By: Teemu, Atte, Alessio, Andrea N.
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 |
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 put the salt on the flame and see that the color it assumes. By the color we determine which anion is in the salt. To understand what cation is in the salt, we calculate it from the atomic mass.
3. Planning
1. Weigh the container with the salt
2. Weigh the empy container.
3. Calculate the weight of the salt
4. Proceed with the flame test
5. See what color the flame becomes
6. Deduce what salt is it.
4. Report
Italian 29th March, repeated 19th April |
Finnish 31.3.2010
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We weight the salt in the container and the weight is 3,54g. The weight of the container is 2,12 g displaced in 5 (elaboration).
We put the salt on the flame and the color resulted yellow-orange.

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empty plastic box= 23,1g.
plastic box + salt= 24,5g.The colour of the flame was yellow.
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Alessio & Andrea
Prof. Alfredo tifi & Eugeni Barbara
5. Elaboration and Interpretation
Italian |
Finnish |
The mass of the salt is (3,54-2,12)g = 1,42 g
With our knowledge, after repetition of the flame test, we deduce that the salt is sodium.
The salt formula is NaX because Na ions are Na+. Now we calculate the proportional number of atoms:
1g (mass of NaCl)
58,44 (atomic mass of NaCl)
= 0,01711156 --> 0,017 (proportional number of atom of calcium sodium, chlorine and halide in our NaX salt)
Now: 0,017 . 22,99 (atomic mass of Na) = 0,393 g (mass of Na)
1,42 - 0,39 = 1,03 g
(mass of X)
1,03
0,017 = 60,59 (atomic mass of X)
The nearest is bromine. Then the salt formula is NaBr.
Our conclusions are that our salt was sodium bromide, NaBr.
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salt mass=1,4 g
metal is sodium
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Common conclusions
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Questions & Answers
Members from both groups are invited to review each other answers first of involving teachers
1. Why are atoms emitting light at the flame?
Because the atom place on the flame will excite and relax itself and take color emits a light impulse according to the energy jump of the transition.
2. How did you recognize the metal element in the salt?
2. Come avete riconosciuto l'elemento metallico presente nel sale?
Risposta 2
With flame test.
3. What measure permitted you to deduce which halogen was in the salt?
3. Quale misura vi ha permesso di dedurre quale alogeno era contenuto nel sale?
Il peso del sale e quale alogeno ho.
4. Does your salt contain more cations or anions?
4. Il vostro sale contiene più cationi o più 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 dell'alogeno, di ioni del metallo o la stessa dei due?
Answer 5
Risposta 5
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